I’m having trouble justifying or verifying the following:
My approach is simple. Draw a new line from a point on the original line to the point above the original line and construct another new line that will be parallel to the original one by making the corresponding angles congruent using the protractor postulate. Uniqueness can be shown through contradiction.
For the perpendicular case, draw a parallel line through the point not on the original line, then draw a line perpendicular to the newly drawn parallel line by using the protractor postulate, which then must also be perpendicular to the original line. Uniqueness again can be shown in many ways.
Lastly I’m having trouble justifying the following problems:
For number $18$, there should be infinitely many lines perpendicular through point $R$, but how do I prove this?
As for $19$ d, again this makes sense I’d imagine it in space, but how does one prove this?
The book you're using is evidently Geometry by Ray C. Jurgensen, Richard G. Brown and John W. Jurgensen.
Your approaches to proving theorems $3$–$8$ and $3$–$9$ are fine, so it's not clear to me why you say you're "having trouble justifying or verifying" them. The only thing missing from your proof outlines are citations to the postulates or previously proved theorems that justify each of their steps. Here are the justifications for each step in the first part your proof outline of theorem $3$-$8$.
First, your choosing a point on the original line is justified by:
Let $\ A,C\ $ be two points on the given line, and $\ O\ $ the given point that lies outside it, and $\ x=m\angle CAO\ $. Drawing the new line through $\ A\ $ and $\ O\ $ is justified by:
Now, to use the protractor postulate at point $\ O\ $, you need to get another point $\ B\ $ on the line through $\ A\ $ and $\ O\ $ such that $\ O\ $ lies between $\ A\ $ and $\ B\ $. You can do this by using the ruler postulate:
Choose a pairing such that $\ A\ $ has coordinate $0$ and $\ O\ $ has coordinate $1$, and let $\ B\ $ be the point with coordinate $2$. Then $\ O\ $ lies between $\ A\ $ and $\ B\ $, so by the protractor postulate there's a ray $\ \vec{OQ}\ $ such that $\ m\angle AOQ=180-x\ $, and $\ m\angle BOQ=m\angle BOA-m\angle AOQ=x\ $ by the angle addition postulate. Now the original line and the line through $\ O\ $ and $\ Q\ $ are cut by the transversal through $\ A\ $ and $\ O\ $, and the corresponding angles $\ \angle CAO\ $ and $\ \angle BOQ\ $ are congruent. Therefore, by postulate $11$, the line through $\ O\ $ and $\ Q\ $ must be parallel to the given line through $\ A\ $ and $\ C\ $.
As you say, you can show uniqueness by contradiction, although I think you can probably also prove it directly. I'll leave it for you to identify the postulates you need for the rest of the proof, and for the proof of theorem $3$-$9$.
I believe the answer to exercise $17$ that the book's authors are looking for is theorem $3$-$9$ itself, rather the angle addition postulate.
I'm not surprised that you're finding it difficult to prove your answers to exercises $18$ and $19$ d, since proving them requires greater familiarity with the techniques needed than you could be expected to have acquired at that stage of the course. Exercise $19$ d is just a restatement of theorem $3$-$10$, given on the immediately preceding page, and I'm sure that the most the book's authors would expect you to do to justify your answer would be to cite that theorem, rather than giving a formally rigorous proof of it. Otherwise, why would exercise $20$ ask you to explain why the result is true in the special case when the three lines are coplanar if you had already proved the general case in exercise $19$ d ?
Likewise, I believe the book's authors would only expect you to base your answer to exercise $18$ on your geometrical intuition rather than by giving a formally rigorous proof.
Nevertheless, since you're inquisitive enough to inquire about the justifications for your answers, I'll give an outline of the proof of theorem $3$-$10$ below.
Proof outline for theorem $\mathbf{3}$-$\mathbf{10}$
We'll need the following definition
from p.$73$, and the following postulates
from p.$23$
Let $\ k, l, n\ $ be three lines with $\ k\parallel l\ $ and $\ l\parallel n\ $. Then, by definition, $\ k\ $ and $\ l\ $ are coplanar and don't intersect, and $\ l\ $ and $\ n\ $ are coplanar and don't intersect. Let $\ p\ $ be the plane containing $\ k\ $ and $\ l\ $, and $\ q\ $ the plane containing $\ l\ $ and $\ n\ $ (that $\ p\ $ and $\ q\ $ are unique follows from postulates $5$ and $7$, and the fact that parallel lines have no point in common. Can you see why?). If it happens that $\ p=q\ $, then all three lines are coplanar, and we have the case for which you're asked to give a justification in exercise $20$, so I'll leave it for you to do that, and just continue on with the case when $\ p\ne q\ $.
Since planes $\ p\ $ and $\ q\ $ intersect (they both contain the line $\ l\ $), postulate $9$ tells us that their intersection is precisely the line $\ l\ $. Therefore line $\ k\ $ cannot intersect plane $\ q\ $ (because it lies in plane $\ p\ $ and doesn't intersect line $\ l\ $, which contains the only points common to planes $\ p\ $ and $\ q\ $) and therefore cannot intersect line $\ n\ $, because line $\ n\ $ lies entirely in plane $\ q\ $.
By postulate $5$ we can choose two points $\ A\ $ and $\ B\ $ on line $\ k\ $ and a third point $\ C\ $ on line $\ n\ $. Postulate $7$ tells us that there's a plane containing the points $\ A, B\ $ and $\ C\ $, so let $\ r\ $ be such a plane. Since point $\ C\ $ lies in plane $\ r\ $ but not in plane $\ p\ $ then $\ r\ne p\ $, and since point $\ A\ $ lies in plane $\ r\ $ but not in plane $\ q\ $ then $\ r\ne q\ $. Since points $\ A\ $ and $\ B\ $ are common to the line $\ k\ $ and the planes $\ p\ $ and $\ r\ $, it follows from postulates $8$ and $9$ that $\ k\ $ is the line of intersection of planes $\ p\ $ and $\ r\ $. By the same reasoning as above, since lines $\ k\ $ and $\ l\ $ are parallel, and both lie in plane $\ p\ $, then line $\ l\ $ cannot intersect the plane $\ r\ $.
Now since $\ r\ne q\ $ it follows from postulate $9$ that $\ r\ $ and $\ q\ $ intersect in a line $\ m\ $, which must contain the point $\ C\ $, since that point lies in both planes $\ r\ $ and $\ q\ $. We have seen above that line $\ l\ $ cannot intersect plane $\ r\ $, so it cannot intersect line $\ m\ $, because that line lies entirely in plane $\ r\ $. Therefore lines $\ l\ $ and $\ m\ $ are parallel, since they are coplanar (both lie in plane $\ q\ $) and do not intersect. But by theorem $3$-$8$ (which you've already proved) there can only be one line through $\ C\ $ which is parallel to $\ l\ $. Therefore $\ m=n\ $, because both $\ m\ $ and $\ n\ $ are parallel to $\ l\ $ and pass through $\ C\ $. Now since both $\ k\ $ and $\ n=m\ $ lie in plane $\ r\ $, then they are coplanar, and we have shown above that they cannot intersect, they are therefore parallel by definition, and the proof is complete. $$ \begin{array}{ccccccccccccccc} &&&&&&&&&&&&&&&&&&&&&& \\ \hline \end{array} $$ As you can see, even though this proof outline glosses over a few points here and there, it's still quite involved. There are some facts about parallel lines and planes that it uses repeatedly and has to justify each time they're used. The proof would have been much easier if these had been proved first, as standalone theorems. The fact that the book doesn't give any such theorems before it introduces theorem $3$-$10$ is another reason why I don't think the authors could (or should) expect you to prove that your answers to exercises $18$ and $19$ d are correct.
The proof of your answer to exercise $18$ looks to me like it's going to be even more involved than that of theorem $3$-$10$, which I expect will actually turn out to be useful in shortening the proof. If you want to have a go at tackling this, here are a few clues on how to go about it.
As you're probably aware, all the lines through $\ R\ $ perpendicular to $\ \stackrel{\longleftrightarrow}{QR}\ $ must lie on a plane perpendicular to $\ \stackrel{\longleftrightarrow}{QR}\ $. So one fundamental part of the proof is to show that you can construct such a plane. One crucial fact you'll need to make use of is the last part of postulate $5$, that "space contains at least four points not all in one plane." Once you've constructed this plane, you can then use the protractor postulate to show that there are an infinite number of rays through $\ R\ $ that lie in the plane. The lines containing these rays are then infinite in number and perpendicular to $\ \stackrel{\longleftrightarrow}{QR}\ $.
I believe that this would be a very ambitious project for you, so you should not be disappointed if you find it too challenging. On the other hand, you'd be entitled to be very chuffed if you do manage to complete it successfully. If you do manage to get started but need some more pointers, I think it would be best for you to ask for them in a separate question.