My answer is different from Xiaoyuan.
(a)
\begin{align*}
Let A &= \begin{bmatrix}
-1 & 1 \\
-1 & -3
\end{bmatrix}\\
~\\
A - \lambda I &= \begin{bmatrix}
-1 - \lambda & 1 \\
-1 & -3 - \lambda
\end{bmatrix}\\
~\\
det(A - \lambda I) &= (\lambda + 3)(\lambda +1) + 1\\
~\\
&= \lambda^{2} + 3\lambda + \lambda + 3 +1\\
~\\
&= (\lambda + 2)^{2} = 0\\
~\\
\lambda_{1} = \lambda_{2} &= -2\\
~\\
For \ \lambda = -2, A - \lambda I &= \begin{bmatrix}
1 & 1 \\
-1 & -1
\end{bmatrix}\\
~\\
Since,\ null(\begin{bmatrix}
1 & 1 \\
-1 & -1
\end{bmatrix}) &= span\{\begin{bmatrix}
1\\
-1 \\
\end{bmatrix}\}
~\\
So \ the \ eigenvector\ v &= \begin{bmatrix}
1\\
-1 \\
\end{bmatrix}\\
~\\
want\ to\ find\ a\ vector\ w, \ where\ &(A - \lambda I)w = v\\
~\\
Then,\ \begin{bmatrix}
1 & 1 \\
-1 & -1
\end{bmatrix}w &= \begin{bmatrix}
1\\
-1 \\
\end{bmatrix}\\
~\\
Then,\ pick\ w &= \begin{bmatrix}
1\\
0 \\
\end{bmatrix}\\
~\\
Thus,\ x(t) &= c_{1}e^{\lambda t}v + c_{2}e^{\lambda t}(w + tv)\\
&= c_{1}e^{-2t}\begin{bmatrix}
1\\
-1 \\
\end{bmatrix} + c_{2}e^{-2t}(\begin{bmatrix}
1\\
0 \\
\end{bmatrix} + t \begin{bmatrix}
1\\
-1 \\
\end{bmatrix})\\
\end{align*}
(b) In the attachment.
(c)
\begin{align*}
Since\ W(x_{1}, x_{2})(t) &= det \begin{bmatrix}
e^{-2t} & e^{-2t} +te^{-2t} \\
-e^{-2t} & -te^{-2t}
\end{bmatrix}\\
~\\
&= -te^{-4t} + e^{-4t} + te^{-4t}\\
~\\
&= e^{-4t} \neq 0\\
~\\
So\ x_{1}(t) \ and \ x_{2}(t) \ form \ a \ fundamental &\ set\ of \ solutions\\
~\\
Then\ the\ fundamenta l\ matrix\ \varphi (t) &= \begin{bmatrix}
e^{-2t} & e^{-2t} +te^{-2t} \\
-e^{-2t} & -te^{-2t}
\end{bmatrix}\\
~\\
Guess\ that\ x(t) &= \varphi (t)\mu(t)\\
~\\
Then\ solve\ that\ \varphi (t)\mu'(t) &= g(t)\\
~\\
\begin{bmatrix}
e^{-2t} & e^{-2t} +te^{-2t} \\
-e^{-2t} & -te^{-2t}
\end{bmatrix} \begin{bmatrix}
\mu_{1}'(t)\\
\mu_{2}'(t) \\
\end{bmatrix} &= \begin{bmatrix}
\frac{e^{-2t}}{t^{2} + 1}\\
0\\
\end{bmatrix}\\
~\\
use\ row\ reduction\ of\ the\ matrix,\ we\ have:\\
~\\
\begin{bmatrix}
e^{-2t} & e^{-2t} +te^{-2t} \\
0 & e^{-2t}
\end{bmatrix} \begin{bmatrix}
\mu_{1}'(t)\\
\mu_{2}'(t) \\
\end{bmatrix} &= \begin{bmatrix}
\frac{e^{-2t}}{t^{2} + 1}\\
\frac{e^{-2t}}{t^{2} + 1}\\
\end{bmatrix}\\
~\\
\mu_{1}'(t) &= \frac{-t}{t^{2} + 1}\\
\mu_{2}'(t) &= \frac{1}{t^{2} + 1}\\
~\\
Then,\ we\ have&:\\
~\\
\mu_{1}(t) &= -\frac{1}{2}\ln(t^{2} + 1) + c_{1}\\
\mu_{2}(t) &= arctan(t) + c_{2}\\
~\\
Thus,\ x(t) &= \varphi (t)\mu(t)\\
~\\
&= \begin{bmatrix}
e^{-2t} & e^{-2t} +te^{-2t} \\
-e^{-2t} & -te^{-2t}
\end{bmatrix} \begin{bmatrix}
-\frac{1}{2}\ln(t^{2} + 1) + c_{1} \\
arctan(t) + c_{2}
\end{bmatrix}\\
~\\
&= \begin{bmatrix}
-\frac{1}{2}e^{-2t}\ln(t^{2} + 1) + (e^{-2t} +te^{-2t})arctan(t)\\
\frac{1}{2}e^{-2t}\ln(t^{2} + 1) - te^{-2t}arctan(t)
\end{bmatrix} + c_{1}e^{-2t}\begin{bmatrix}
1\\
-1 \\
\end{bmatrix} + c_{2}e^{-2t}(\begin{bmatrix}
1\\
0 \\
\end{bmatrix} + t \begin{bmatrix}
1\\
-1 \\
\end{bmatrix})\\
~\\
&= e^{-2t}\ln(t^{2} + 1) \begin{bmatrix}
-\frac{1}{2}\\
\frac{1}{2}\\
\end{bmatrix} + te^{-2t}arctan(t) \begin{bmatrix}
1\\
-1 \\
\end{bmatrix}) + e^{-2t}arctan(t) \begin{bmatrix}
1\\
0 \\
\end{bmatrix} + c_{1}e^{-2t}\begin{bmatrix}
1\\
-1 \\
\end{bmatrix} + c_{2}e^{-2t}(\begin{bmatrix}
1\\
0 \\
\end{bmatrix} + t \begin{bmatrix}
1\\
-1 \\
\end{bmatrix})\\
~\\
Sine\ x(0) &= \begin{bmatrix}
0\\
0\\
\end{bmatrix}\\
~\\
Then,\ c_{1} + c_{2} &= 0 \ and\ -c_{1} = 0\\
~\\
Thus,\ c_{1} &= c_{2} = 0\\
~\\
Therefore,\ x(t) &= e^{-2t}\ln(t^{2} + 1) \begin{bmatrix}
-\frac{1}{2}\\
\frac{1}{2}\\
\end{bmatrix} + te^{-2t}arctan(t) \begin{bmatrix}
1\\
-1 \\
\end{bmatrix}) + e^{-2t}arctan(t) \begin{bmatrix}
1\\
0 \\
\end{bmatrix}\\
\end{align*}
