Let $g$ be the line with equation $y=1$ and let $A=(0,2)$ and $B=(2,0)$.
Show that there is exactly one glide reflection $\kappa$ with fixed line $g$ that maps $A$ on $B$.
Show that there is exactly one rotation $\delta$ with fixed line $g$ that maps $A$ on $B$.
To show the existence of $\kappa$ we have to show that the distance of $A$ to $g$ and of $B$ to $g$ is the same, or not?
But how can we show the uniqueness?
A glide reflection can be expressed as a composition of a reflection about a line and a translation parallel to that line.
A glide reflection composed of a reflection about a line $L$ followed by translation parallel to $L$ will always map $L$ to $L$, that is, $L$ is a fixed line of the glide reflection. If the translation is non-trivial (that is, it is a translation through a non-zero distance), then this line $L$ is the only fixed line of the glide reflection.
Therefore when we are told $g$ is a fixed line of glide reflection $\kappa,$ we know that it must be possible to describe $\kappa$ as a reflection about $g$ followed by a translation parallel to $g.$
What happens when we reflect the point $A$ about the line $g$? Is there more than one possible result?
Find the image of $A$ when $A$ is reflected through $g.$ Call the image $A'$. What translation (if any) parallel to $g$ takes $A'$ to $B$?
If you work out the composition of $\kappa$ from a reflection about $g$ and a translation parallel to $g,$ you should be able to show that this can be done in only one way, that is, there is only one possible reflection and one possible translation that could compose to $\kappa.$
It is true that it would be impossible to find any such glide reflection $\kappa$ if $A$ and $B$ were at different distances from $g.$ But you don't actually need to prove explicitly that they are. Finding a translation from $A'$ to $B$ parallel to $g$ will prove it implicitly.
For part 2 of the question:
Rotations in the plane are rotations around a single point. It is impossible to rotate a plane about a line in the plane.
In fact, if an isometry of the plane maps $P$ to $P$ and maps $Q$ to $Q$, where $P$ and $Q$ are distinct points, the isometry is either the identity or is a reflection through the line $PQ.$
But my understanding is that an object is "fixed" by an isometry of a plane if the image of the object is the object itself; it is not necessary that every point of the object is a fixed point of the isometry. This is how the line $g$ is able to be a "fixed line" of the glide reflection in part 1, which has no fixed points at all.
So if we say a rotation has a fixed line then we mean every point on that line is mapped to itself or to another point on the line.
The line $g$ obviously is fixed by the identity transformation, which can also be considered a rotation around some point by a multiple of $2\pi$ radians.
The line $g$ also is fixed if the center of rotation is on $g$ and the angle of rotation is $\pi$ radians. A rotation by $\pi$ radians keeps the center of the rotation fixed and swaps every other point of $g$ with a point of $g$ on the opposite side of the center of rotation.
Suppose the rotation is not by a whole multiple of $\pi$ or is around some point not on the line $g$ (and not the identity transformation). In each case you can show some point on the line $g$ that is mapped to a point not on $g.$
So now you know the angle of rotation and you can limit the location of the center to just a few possible places. In this particular question, however, given that the rotation $\delta$ maps a known point $A$ to a known point $B$, it is sufficient to know the angle of rotation in order to find the center.
If you want to do it with matrices and vectors I suppose you could write a rotation matrix with unknown angle $\theta$ and a translation vector with unknown coordinates, plug in the coordinates of $A$ and $B,$ and solve for the unknowns. The geometric explanation is so compelling, however, that I am not inclined to work through all that algebra.