I'm wondering if there is also another easy way of solving question C. In the book, they use $A = PDP^{-1}$ but the only method I know is by drawing the transformation with vectors $ e_1$ and $e_2$ and then look at the result and that would be the standard matrix. I tried this method and it did not work for me. I took into account that the vectors only turned so I kept the total length of the vector equal by using unit vectors of the transformation.
On
You can do some geometry:
Denote $(x,y) =T(e_1)$. The line $e_1 T(e_1)$ is orthogonal to the directing vector $(2,1)$, i.e. $$2(x-1)+1\cdot y=0 $$ Further, $T(e_1)$ has norm $1$, so $$x^2+y^2=x^2+4(x-1)^2=1\iff 5x^2-8x+3=0.$$ This quadratic equation has two roots: $x=1$, corresponding to vector $e_1$, and $\frac35$, whence $y=\frac45$, corresponding to $T(e_1)$.
Can you do the same for $T(e_2)$ and deduce the matrix $A$?
Having the two eigenvectors, the transformation matrix is $$P=\begin{pmatrix}2 & -1 \\1 & 2\end{pmatrix}.$$ whereas the diagonal matrix would be $$D=\begin{pmatrix}1 & 0 \\0 & -1\end{pmatrix}$$ Using these, you can calculate $A$.
If you want to use your approach, you ought to calculate the image of the two basic vectors $e_1,e_2$. One can calculate that using basic geometry, to obtain $$T(e_1)=T\begin{pmatrix} 1 \\ 0\end{pmatrix}=\begin{pmatrix} 0.6 \\ 0.8\end{pmatrix}$$ and $$T(e_2)=T\begin{pmatrix} 0 \\ 1\end{pmatrix}=\begin{pmatrix} 0.8 \\ -0.6\end{pmatrix}$$ thus $$A=\begin{pmatrix} 0.6 & 0.8 \\ 0.8 & -0.6\end{pmatrix}$$