$\bf(1)$ Let $\mathbb{R}^{2\times2}$ denote the vector space of all $2\times2$ matrices with real number entries. Set $A=\pmatrix{1&2\\-2&-4}$. Define a linear transformation $T:\mathbb{R}^{2\times2}\longrightarrow\mathbb{R}^{2\times2}$ by $T(B)=AB$ for any $B\in\mathbb{R}^{2\times2}$.
$\quad(a)$ Compute the $\alpha$-matrix $[T]_\alpha$ for $\mathrm T$, where $\alpha$ denotes the following ordered basis for $\mathbb{R}^{2\times2}:$ $$\pmatrix{0&0\\1&1},\pmatrix{1&0\\0&0},\pmatrix{0&0\\-1&1},\pmatrix{0&1\\0&1}.$$
$\qquad\bf Solution:$ $[T]_\alpha=\pmatrix{-5 & -1 & -1 & -\tfrac92 \\
2 & 1 & -2 & 0 \\
-1 & 1 & -5 & -\tfrac92 \\
2 & 0 & 2 & 3}$.
$\quad(b)$ Is $\rm T$ an isomorphism?
$\bf Solution:$ No $\rm T$ is not an isomorphism. $\rm T$ can not be "onto" since $\operatorname{rank}([T]_\alpha)<4$. Another reason $\rm T$ can not be "onto" is that the $2\times2$ identity matrix $I_2$ is not a value of $\rm T$: if it were a value then we would have that $AB=I_2$ for some $2\times2$ matrix $\rm B$, which would imply that $A$ is invertible - which is not true.
$\quad(c)$ Compute the $\alpha$-coordinates of $\pmatrix{\pi & 3 \\ \sqrt{2}&0}$.
$\qquad\bf Solution:$ $\pmatrix{\tfrac{\sqrt{2}-3}2 \\ \pi \\ \tfrac{-\sqrt{2}-3}2 \\ 3}$
I'm having trouble with parts A and C.
$$T\begin{pmatrix}0&0\\1&1\end{pmatrix}:=\begin{pmatrix}\;\;2&\;\;2\\-4&-4\end{pmatrix}=\color{red}{-5}\begin{pmatrix}0&0\\1&1\end{pmatrix}+\color{red}2\begin{pmatrix}1&0\\0&0\end{pmatrix}+\color{red}{(-1)}\begin{pmatrix}\;\;0&0\\-1&1\end{pmatrix}+\color{red}2\begin{pmatrix}0&1\\0&1\end{pmatrix}$$
and that's how you get the wanted matrix's first column. Continue from here applying $\;T\;$ to the other members of that basis and writing the result as a linear combination of that same basis
For (c): write that vector as a linear combination of the basis:
$$\begin{pmatrix}\pi&3\\\sqrt2&0\end{pmatrix}=a\begin{pmatrix}0&0\\1&1\end{pmatrix}+b\begin{pmatrix}1&0\\0&0\end{pmatrix}+c\begin{pmatrix}\;\;0&0\\-1&1\end{pmatrix}+d\begin{pmatrix}0&1\\0&1\end{pmatrix}$$
For example,. it must clearly be that $\;b=\pi\;$ (why?), etc.