You have an eigenvalue $\lambda$ and its eigenvector $v_1$. So one of your solutions will be
$$ x(t) = e^{\lambda t} v_1$$
As you've noticed however, since you only have two eigenvalues (each with one eigenvector), you only have two solutions total, and you need four to form a fundamental solution set. For each eigenvalue $\lambda$, you will calculate what's called a generalized eigenvector $v_2$, which is the solution to
$$ (A - \lambda I)v_2 = v_1, \quad \text{where } (A - \lambda I)v_1 = 0;$$
in other words, $v_1$ was the first eigenvector. Then this contributes a new solution
$$ x(t) = e^{\lambda t} v_2 + te^{\lambda t} v_1 $$
Now you have two linearly independent solutions corresponding to one eigenvalue. Now repeat the process for the second eigenvalue to get all four elements of your fundamental solution set.
Note: What I'm calling $v_2$ is NOT the same as what you're calling $v_2$ in your question.
The solution to the vector differential equation $\mathbf y'(x)=A\mathbf y(x)$ is, not surprisingly, $e^{xA}\mathbf C$, where $\mathbf C$ is a vector of constants determined by boundary conditions. The exponential of a matrix is defined via a power series, but in practice one doesn’t use that to compute it.
If $A$ is diagonalizable, it can be decomposed as $B\Lambda B^{-1}$, where $B=\begin{bmatrix}\mathbf b_1,\cdots,\mathbf b_n\end{bmatrix}$ with eigenvectors of $A$ as its columns and $\Lambda=\operatorname{diag}(\lambda_1,\dots,\lambda_n)$ with $\lambda_j$ the eigenvalues corresponding to $\mathbf b_j$. Just as $A^k=B\Lambda^kB^{-1}=B\operatorname{diag}(\lambda_1^k,\dots,\lambda_n^k)B^{-1}$, so, too $e^{xA}=Be^{x\Lambda}B^{-1}=B\operatorname{diag}(e^{\lambda_1x},\dots,e^{\lambda_nx})B^{-1}$. So, the solution to the equation is $$e^{xA}\mathbf C=Be^{x\Lambda}B^{-1}\mathbf C.$$ Since $B$ has full rank, $B^{-1}\mathbf C$ is also a vector of arbitrary constants, so we can expand the above expression as $$C_1e^{\lambda_1x}\mathbf b_1+\cdots+C_ne^{\lambda_nx}\mathbf b^n.\tag1$$
In this problem, $$A=\begin{bmatrix}-2&1&-2\\1&-2&2\\3&-3&5\end{bmatrix}.$$ Its eigenvalues can be found to be $3$, $-1$ and $-1$. We compute the kernel of $3I-A$ via row-reduction: $$\begin{bmatrix}5&-1&2\\-1&5&-2\\-3&3&-2\end{bmatrix}\to\begin{bmatrix}1&0&\frac13\\0&1&-\frac13\\0&0&0\end{bmatrix}$$ so an eigenvector of $3$ is $(-1,1,3)^T$. Moving to $A+I$, $$\begin{bmatrix}-1&1&-2\\1&-1&2\\3&-3&6\end{bmatrix}\to\begin{bmatrix}1&-1&2\\0&0&0\\0&0&0\end{bmatrix}$$ from which we get $(1,1,0)^T$ and $(-2,0,1)^T$ as linearly independent eigenvectors. Plugging these values into (1), the solution to the equation is therefore $$C_1e^{3x}\begin{bmatrix}-1\\1\\3\end{bmatrix}+C_2e^{-x}\begin{bmatrix}1\\1\\0\end{bmatrix}+C_3e^{-x}\begin{bmatrix}-2\\0\\1\end{bmatrix}=\begin{bmatrix}-C_1e^{3x}+(C_2-2C_3)e^{-x}\\C_1e^{3x}+C_2e^{-x}\\3C_1e^{3x}+C_3e^{-x}\end{bmatrix}.$$
Best Answer
I am going to rewrite the system as
$$\tag 1 x'' + y' + 3 x = e^{-t} \\ y'' - 4 x' + 3 y = \sin 2t $$
Using the first equation, we have
$$\tag 2 y' = -x'' - 3 x + e^{-t} \\ y'' = -x''' - 3 x' - e^{-t} \\ y''' = - x'''' - 3 x'' + e^{-t}$$
Taking the derivative of the second equation
$$\tag 3 y''' - 4 x'' + 3 y' = 2 \cos 2 t$$
Substituting $(2)$ into $(3)$
$$(- x'''' - 3 x'' + e^{-t}) - 4 x'' + 3 (-x'' - 3 x + e^{-t}) = 2 \cos 2t $$
This simplifies to
$$-x'''' - 10 x'' - 9x = 2 \cos 2 t - 4 e^{-t} $$
Solving this using your favorite method
$$x(t)= c_1 \cos 3t +c_2 \sin 3t+c_3 \cos t + c_4 \sin t+\dfrac{e^{-t}}{5}+\dfrac{2}{15} \cos 2t$$
Now, using the first equation and $x(t)$, we need to solve
$$\tag 4 y' = -x'' - 3 x + e^{-t} = 6 c_1 \cos 3t+6 c_2 \sin 3t-2 c_3 \cos t -2 c_4 \sin t+\dfrac{e^{-t}}{5}+\dfrac{2}{15} \cos 2t$$
By integrating $(4)$, we get
$$y(t) = 2 c_1 \sin 3t-2 c_2 \cos 3t-2 c_3 \sin t+2 c_4 \cos t-\dfrac{e^{-t}}{5}+\dfrac{1}{15} \sin 2t$$
You can substitute $(1)$ and $(2)$ into the original system and verify they are solutions.