According to the figure here, how to find the standard matrix for H(θ) without using formulas?
Another question is that for example 6 and 7, the question is exactly the same, but why the answers are different?
What's the difference between the two questions ? And how do I know when two use formula 16 and when to use formula 18?
The statement of the problem in Example 7 is mistaken (I believe the authors have copied & pasted the one in Example 6 without making the necessary changes). The correct version should read (the modifications are in boldface):
When you are required to compute orthogonal projections on a line, use $P_\theta$ from formula 16 (easy to remember: $P$ comes from "projection"). When required to compute reflections about a line, use $H_\theta$ from formula 18 (and no, $H$ doesn't come from anything this time...).
Concerning your first question, the reflection of $\Bbb x$ about the line of slope $\tan \theta$ will obviously depend on $\theta$, dependence encoded in formulae. You cannot not have formulae...
Edit: To answer the question in the comment, note that $(1,5) = 1 \cdot e_1 + 5 \cdot e_2$, therefore $H_\theta (1,5) = 1 \cdot H_\theta e_1 + 5 \cdot H_\theta e_2$. Therefore, once we find out how the vectors $e_1$ and $e_2$ reflect about the line $L$, we are done. Fortunately, we can do this without complicated formulae; take a pencil and a sheet of paper and draw while reading this answer. Call the reflections of $e_1$ and $e_2$ respectively $r_1$ and $r_2$.
The $e_1$ vector is below $L$, forming an angle of $30 ^\circ$ with it, therefore its reflection will be above $L$, at the same angle, so $r_1$ will be at an angle of $30 ^\circ + 30 ^\circ = 60 ^\circ$ above the $x$ axis. Now, look at the right triangle formed by $0$ (the origin), the tip of the reflection if $r_1$, and the projection of $r_1$ on the $x$ axis; the length of the horizontal cathetus (i.e. leg) will give the $x$ coordinate of $r_1$, and the length of the vertical cathetus will give the $y$ coordinate of $r_1$. But the length of the horizontal cathetus is the length of the hypothenuse times $\cos 60 ^\circ$; since the length of the hypothenuse is the length of $r_1$, which is the length of $e_1$, which is $1$, the length of the horizontal cathetus will be $\cos 60 ^\circ = \frac 1 2$. Similarly, the $y$ cordinate of $r_1$ will be the length of the other cathetus, i.e. $\sin 60 ^\circ = \frac {\sqrt 3} 2$. Therefore, $r_1 = (\frac 1 2, \frac {\sqrt 3} 2)$.
The same discussion now for $e_2$: it is above $L$, making $60 ^\circ$ with it, therefore $r_2$ will be below $L$, at the same angle, which means that it will be below the $x$ axis, making $30 ^\circ$ with it. Again, investigating the right triangle formed (under the $x$ axis) by $0$, the tip of $r_2$ and the projection of $r_2$ on the $x$ axis, you will get $r_2 = (\frac {\sqrt 3} 2, - \frac 1 2)$ (the minus comes from being below the $x$ axis, where $y$ is negative).
Putting these together, you get $H_\theta (1,5) = 1 \cdot (\frac 1 2, \frac {\sqrt 3} 2) + 5 \cdot (\frac {\sqrt 3} 2, - \frac 1 2) = (\frac {5 \sqrt 3 + 1} 2, \frac {\sqrt 3 -5} 2)$ in a purely graphical way, without matrix multiplications (if this is what you meant by "witout formulae"). Please note, though, that some formulae are still necessary, you cannot expect to do mathematics without them save for in some extremely simple situations.