11. Given two sequences $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ satisfy
$$
\begin{array}{l}
a_{1}=2, b_{1}=1, \\
a_{n+1}=5 a_{n}+3 b_{n}+7\left(n \in \mathbf{Z}_{+}\right), \\
b_{n+1}=3 a_{n}+5 b_{n}\left(n \in \mathbf{Z}_{+}\right) .
\end{array}
$$

Then the general term formula for $\left\{a_{n}\right\}$ is $a_{n}=$ $\qquad$
11. $2^{3 n-2}+2^{n+1}-4$.
(1) + (2) +1 gets
$$
a_{n+1}+b_{n+1}+1=8\left(a_{n}+b_{n}+1\right) \text {, }
$$

thus $a_{n}+b_{n}+1=8^{n-1}\left(a_{1}+b_{1}+1\right)=2^{3 n-1}$;
(1) - (2) +7 gets
$$
a_{n+1}-b_{n+1}+7=2\left(a_{n}-b_{n}+7\right) \text {, }
$$

thus $a_{n}-b_{n}+7=2^{n-1}\left(a_{1}-b_{1}+7\right)=2^{n+2}$.
Therefore, $a_{n}=\frac{2^{3 n-1}+2^{n+2}-8}{2}=2^{3 n-2}+2^{n+1}-4$.