We consider the isometries of $\Bbb R^2$. Let $\varphi$ be the rotation of $90^\circ$ (counterclockwise) around the point $\begin{pmatrix}3 \\ 5\end{pmatrix}$ and let $\psi$ be the reflection about the line that is parallel to the $x$-axis through the point $\begin{pmatrix}0 \\ 1\end{pmatrix}$.
- Determine $v\in\Bbb R^2,\ a\in O_2$ s. t. $\varphi=\tau_v\circ\varphi_a$.
- Determine $w\in\Bbb R^2$, $b\in O_2$ such that $\psi=\tau_w\circ\varphi_b$.
- Determine $u\in\Bbb R^2,\ c\in O_2$ s. t.$\psi\circ\varphi=\tau_u\circ\varphi_c$.
$$$$
For the first one we have that: $$(\tau_v\circ\varphi_a)(x)=\tau_v(\varphi_a(x))=\tau_v(ax)=ax+v$$ And this has to be equal to the vector $x$ rotated by $90^\circ$ counterclockwise around the point $\begin{pmatrix}3 \\ 5\end{pmatrix}$, right?
So do we have to use the matrix for the map of rotation?
I assume that $O_2$ refers to the set of $2 \times 2$ orthogonal matrices, so that $a$ for instance is meant to represent a $2 \times 2$ matrix. Here's an approach that avoids using homogeneous coordinates, contrary to the suggestions of my comment.
Let $r_{\theta}$ denote the matrix corresponding to the counterclockwise rotation by angle $\theta$. Note (or show) that the rotation $\phi(x)$ satisfies $$ \phi(x) = r_{90^\circ}(x - (3,5)^T) + (3,5)^T. $$ Expand the multiplication above to obtain an expression of the form $ax + v$, which gives you the answer to the first part.
For part 2, note similarly that if $f_{(a,b)}$ denotes the reflection across the line through $0$ in the direction of $(a,b)^T$, then we can write $$ \psi(x) = f_{(1,0)}(x - (0,1)^T) + (0,1)^T. $$