System of Distributional Differential Equations 
Here are the problem and my attempt to the solution. Is it correct?
 A: Let
$$
A = \begin{pmatrix}2 & -1 \\ 3 & -2\end{pmatrix},
\quad
y = \begin{pmatrix}y_1 \\ y_2\end{pmatrix},
\quad
b = \begin{pmatrix}0 \\ 2\end{pmatrix} \delta(x).
$$
The equation can then be written 
$y' = Ay + b.$
This can be rewritten as
$y' - Ay = b.$
After multiplication from the left with $e^{-Ax}$ we get
$e^{-Ax}y' - e^{-Ax}Ay = e^{-Ax}b.$
The left hand side can now be written as $(e^{-A}y)'$ giving
$(e^{-Ax}y)' = e^{-Ax}b,$
i.e. $e^{-Ax}y = \int e^{-Ax}b + c,$
where $c$ is a constant vector and $\int$ denotes taking antiderivative.
Thus the solution is given by
$$y = e^{Ax} \int e^{-Ax}b + e^{Ax} c.$$
Now $A^2 = I$ (check it!) so 
$$
e^{Ax} 
= \sum_{n=0}^{\infty} \frac{1}{n!} (Ax)^n
= \sum_{n\text{ even}} \frac{1}{n!} x^n I + \sum_{n\text{ odd}} \frac{1}{n!} x^n A
= (\cosh x) I + (\sinh x) A
$$
and
$$
e^{-Ax} = (\cosh x) I - (\sinh x) A
.
$$
This makes
$$\begin{align}
e^{-Ax} b
&= \left( (\cosh x) I - (\sinh x) A \right) b \\
&= 
\begin{pmatrix}
\cosh x - 2 \sinh x & \sinh x \\
-3 \sinh x & \cosh x + 2 \sinh x
\end{pmatrix}
\begin{pmatrix}
0 \\
2
\end{pmatrix} \delta(x) \\
&=
\begin{pmatrix}
2 \sinh x \\
2 \cosh x + 4 \sinh x
\end{pmatrix}
\delta(x) \\
&=
\begin{pmatrix}
0 \\
2
\end{pmatrix}
\delta(x),
\end{align}$$
where the last identity holds since $f(x) \, \delta(x) = f(0) \, \delta(x).$
Now,
$$
\int e^{-Ax} b = \begin{pmatrix}0 \\ 2\end{pmatrix} H(x),
$$
where $H(x)$ is the Heaviside function.
This gives
$$
e^{Ax} \int e^{-Ax} b
= \left( (\cosh x) I + (\sinh x) A \right)
\begin{pmatrix}0 \\ 2\end{pmatrix} H(x)
= \begin{pmatrix}-2 \sinh x \\ 2 \cosh x - 4 \sinh x\end{pmatrix} H(x).
$$
Thus the general solution is
$$
y(x) = 
\begin{pmatrix}
-2 \sinh x\\ 
2 \cosh x - 4 \sinh x
\end{pmatrix} H(x)
+ \begin{pmatrix}
c_1 \cosh x + 2 c_1 \sinh x - c_2 \sinh x \\
c_2 \cosh x + 3 c_1 \sinh x - 2 c_2 \sinh x
\end{pmatrix},
$$
where $c_1$ and $c_2$ are constants.

Solving the equation using Fourier transform, just like you I get
$$
\begin{pmatrix}\hat{y}_1 \\ \hat{y}_2\end{pmatrix}
= \frac{-1}{1+\xi^2} \begin{pmatrix}i\xi+2 & -1 \\ 3 & i\xi-2\end{pmatrix}
\begin{pmatrix}0 \\ 2\end{pmatrix}
$$
but then you make some mistakes. First you make a sign error. I get
$$
\begin{pmatrix}\hat{y}_1 \\ \hat{y}_2\end{pmatrix}
= \frac{2}{1+\xi^2} \begin{pmatrix}1 \\ 2-i\xi\end{pmatrix}.
$$
The right hand side is the Fourier transform of the convolution
$$
e^{-|x|} * \begin{pmatrix}\delta \\ 2\delta-\delta' \end{pmatrix}
= \begin{pmatrix}e^{-|x|} * \delta \\ e^{-|x|}*(2\delta-\delta') \end{pmatrix}
= \begin{pmatrix}e^{-|x|} \\ 2 e^{-|x|} - (e^{-|x|})' \end{pmatrix} \\
= \begin{pmatrix}e^{-|x|} \\ 2 e^{-|x|} + \operatorname{sign}(x) e^{-|x|} \end{pmatrix} 
= e^{-|x|} \begin{pmatrix}1 \\ 2 + \operatorname{sign}(x) \end{pmatrix}.
$$
Thus,
$$
\begin{pmatrix}{y}_1 \\ {y}_2\end{pmatrix}
= e^{-|x|} \begin{pmatrix}1 \\ 2 + \operatorname{sign}(x) \end{pmatrix}
+ \text{solutions to the homogeneous equation}.
$$
I should add something about the solutions to the homogeneous equation and also show that the two solutions are equal.
