= Solution
The first and third equations form a constant-coefficient <linear system of differential equations>
$$
\frac d{dt}\binom{x}{z}
=\begin{pmatrix}-1&3\\3&-1\end{pmatrix}\binom{x}{z}.
$$
The <eigenvectors> $(1,1)$ and $(1,-1)$ have <eigenvalues> $2$ and $-4$. Applying the initial data gives
$$
x(t)=\frac{e^{2t}-e^{-4t}}2,
\qquad
z(t)=\frac{e^{2t}+e^{-4t}}2.
$$
The remaining equation becomes
$$
y'-2y=-3e^{-4t}+\cos t-2\sin t.
$$
An <integrating factor>, or the usual exponential and trigonometric particular solutions, gives
$$
\boxed{x(t)=\frac{e^{2t}-e^{-4t}}2,
\quad y(t)=\sin t+\frac{e^{-4t}-e^{2t}}2,
\quad z(t)=\frac{e^{2t}+e^{-4t}}2}.
$$
These values satisfy all three initial conditions.
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