$\underline{\textbf{Solution:}}$ $\\$
$\textbf{(a)}$ $\\$
Find eigenvalues by $\det(A-\lambda I_2)=0$:
$$\begin{array}{|c c|}-1-\lambda&1\\0&-3-\lambda\end{array}=0 \implies \lambda^2+4\lambda+3=0\implies \cases{\lambda_1=-1\\ \lambda_2=-3}$$
Find eigenvectors by $(A-\lambda I_2)\mathbf{x}=\boldsymbol 0$: $\\$
When $\lambda=-1$, eigenvector $\boldsymbol{\xi}^{(1)}=\begin{pmatrix}1\\0\end{pmatrix}$. $\\$
When $\lambda=-3$, eigenvector $\boldsymbol{\xi}^{(2)}=\begin{pmatrix}1\\-2\end{pmatrix}$. $\\$
(Few steps omitted, since "Syed_Hasnain" got part(a) right already.)$\\$
Therefore, the general solution is $$\mathbf{x}(t)=c_1\begin{pmatrix}1\\0\end{pmatrix}e^{-t}+c_2\begin{pmatrix}1\\-2\end{pmatrix}e^{-3t}$$
$\textbf{(b)}$ $\\$
$$W[\mathbf{x}^{(1)},\mathbf{x}^{(2)}](t)=\begin{array}{|c c|}e^{-t}&e^{-3t}\\0&-2e^{-3t}\end{array}=-2e^{-4t}\neq 0$$
Thus, $\mathbf{x}^{(1)}(t)$ and $\mathbf{x}^{(2)}(t)$ form a fundamental set of solutions. $\\$
Hence the fundamental matrix $$\boldsymbol\Psi(t)=\begin{pmatrix}e^{-t}&e^{-3t}\\0&-2e^{-3t}\end{pmatrix}$$
For the non-homogeneous system, we have the general solution$$\mathbf{x}=\boldsymbol\Psi(t)\mathbf{u}(t)$$
Since we know $$\boldsymbol\Psi(t)\mathbf{u}'(t)=\mathbf{g}(t)$$
$$\begin{pmatrix}e^{-t}&e^{-3t}\\0&-2e^{-3t}\end{pmatrix}\begin{pmatrix}u_1'\\u_2'\end{pmatrix}=\begin{pmatrix}{12\over e^t+1}\\{12\over e^t+1}\end{pmatrix}$$
By row reduction: $$\begin{pmatrix}e^{-t}&0\\0&-2e^{-3t}\end{pmatrix}\begin{pmatrix}u_1'\\u_2'\end{pmatrix}=\begin{pmatrix}{18\over e^t+1}\\{12\over e^t+1}\end{pmatrix}$$
Hence $$\cases{u_1'={18e^{t}\over e^t+1}\\u_2'={-6e^{t}\over e^t+1}} \implies \cases{u_1(t)=\int{{18e^{t}\over e^t+1}dt}=18\ln(e^t+1)+c_1\\u_2(t)=\int{{-6e^{t}\over e^t+1}dt}=-6\ln(e^t+1)-3e^{-2t}+6e^t+c_2}$$
Thus $$\begin{align}\mathbf{x}&=\boldsymbol\Psi(t)\mathbf{u}(t)\\&=c_1\begin{pmatrix}1\\0\end{pmatrix}e^{-t}+c_2\begin{pmatrix}1\\-2\end{pmatrix}e^{-3t}+\begin{pmatrix}-3\\6\end{pmatrix}e^{-5t}+\begin{pmatrix}6\\-12\end{pmatrix}e^{-2t}+\ln(e^t+1)\begin{pmatrix}-6\\12\end{pmatrix}e^{-3t}+\ln(e^t+1)\begin{pmatrix}18\\0\end{pmatrix}e^{-t}\end{align}$$
Apply the given initial condition:$$\mathbf{x}(0)=\begin{pmatrix}12\ln2+c_1+c_2+3\\12\ln2-6-2c_2\end{pmatrix}=\begin{pmatrix}1-3\ln2\\-3-3\ln2\end{pmatrix}\implies \cases{c_1=-{1\over2}-{45\over2}\ln2\\c_2=-{3\over2}+{15\over2}\ln2}$$
Therefore, the general solution for the IVP:
$$\mathbf{x}=(-{1\over2}-{45\over2}\ln2)\begin{pmatrix}1\\0\end{pmatrix}e^{-t}+(-{3\over2}+{15\over2}\ln2)\begin{pmatrix}1\\-2\end{pmatrix}e^{-3t}+\begin{pmatrix}-3\\6\end{pmatrix}e^{-5t}+\begin{pmatrix}6\\-12\end{pmatrix}e^{-2t}+\ln(e^t+1)\begin{pmatrix}-6\\12\end{pmatrix}e^{-3t}+\ln(e^t+1)\begin{pmatrix}18\\0\end{pmatrix}e^{-t}$$
