If $\cal B=\{ {\bf v},{\bf w}\}$ is an ordered basis of $\Bbb R^2$, then the matrix representation, $M$, of $A:\Bbb R^2\rightarrow\Bbb R^2$ with respect to this basis (I assume you want both the domain and range to have basis $\cal B$) is the $2\times 2$ matrix that has as its first column the coordinates of $A{\bf v}$ with respect to $\cal B$ and as its second column the coordinates of $A{\bf w}$ with respect to $\cal B$.
What does this mean? Well, if you write a vector ${\bf x}$ in terms of this basis ${\bf x}= c_1{\bf v}+c_2{\bf w},$ then, setting, $[{\bf x}]_{\cal B}=[{c_1\atop c_2}]$ $ \tag {1}[A {\bf x } ]_{\cal B} = M[{\bf x }]_{\cal B}. $
That is, for $\bf x$ written in the standard basis, the coordinates of $A{\bf x}$ with respect to $\cal B$ are given by the product of the matrix $M$ with the coordinate matrix of $\bf x$ with respect to $\cal B$.
For part 1.:
The matrix representation of $A$ is easily found, since you were told what $A\bigl((1,1)\bigr)$ and $A\bigl((1,0)\bigr)$ were.
We need to write $(2,1)$ and $(0,3)$ in terms of the basis $\cal B=\{(1,1),(1,0)\}$.
$(2,1)= 1 (1,1)+1(1,0)$ and $ (0,3)= 3(1,1)-3(1,0) $ The matrix $M$ is $M=\Bigl[\, \underbrace{1\atop 1}_{ [A(1,1)]_{\cal B} } \ \underbrace{3\atop -3}_{ [A(1,0)]_{\cal B} }\,\Bigr].$
For part 2.:
You need to write $(3,2)$ in terms of $\cal B$: $ (3,2)=2(1,1)+1(1,0). $
Using the matrix representation of $A$, $[A\bigl((3,2 )\bigr)]_{\cal B}=\Bigl [\, {1\atop 1}\ {3\atop -3}\,\Bigr ]\Bigl[ {2\atop 1}\Bigr]=\Bigl[{5\atop -1} \Bigr]. $
This gives the coordinates of $A((3,2))$ with respect to $\cal B$, so $ A((3,2))= 5(1,1)+(-1)(1,0)=(4,5). $
For part 3.: Let \cal B'=\{(x_1,x_2), (0,3)\}
You know the matrix $W= \Bigl[\,{-6\atop 3}\ {-6\atop4}\,\Bigr ] $ is the matrix representation of $A$ with respect to \cal B'.
The second column of $W$ is [ A\bigl((0,3)\bigr)]_{\cal B'}.
So, $\tag{2}A((0,3))=-6(x_1,x_2)+4(0,3)=(-6x_1, -6x_2+12).$
But you can compute $A\bigl((0,3)\bigr)$ using the matrix representation from part 1.
We find $[A\bigl((0,3)\bigr)]_{\cal B}$ first. Towards this end, we write $(0,3)$ in terms of the basis $\cal B$ first. Solve:
$ (0,3) = c_1(1,1)+c_2(1,0) $ to obtain $ \eqalign{ c_1&=3\cr c_2&=-3. } $ Then: $ [A\bigl((0,3)\bigr)]_{\cal B}=\Bigl [\, {1\atop 1}\ {3\atop -3}\,\Bigr ]\Bigl[ {3\atop -3}\Bigr]=\Bigl[{ -6\atop 12}\Bigr]. $
So, the coordinates of $A((0,3))$ with respect to $\cal B$ are $(-6,12)$. So, $ \tag{3}A((0,3))=-6(1,1)+12(1,0)= (6, -6) $ Comparing equations (2) and (3) gives $ \eqalign{ 6&=-6x_1\cr -6&=-6x_2+12 } $
This gives $x_1=-1$ and $x_2=3$.