9. The sequences $\left\{a_{n}\right\},\left\{b_{n}\right\}$ satisfy $a_{1}=2, b_{1}=1,\left\{\begin{array}{l}a_{n+1}=5 a_{n}+3 b_{n}+7 \\ b_{n+1}=3 a_{n}+5 b_{n}\end{array} \quad\left(n \in \mathbf{N}^{*}\right)\right.$, then their general terms are $a_{n}=$ $\qquad$ $b_{n}=$ $\qquad$ .
9. $a_{n}=2^{3 n-2}+2^{n+1}-4, b_{n}=2^{3 n-2}-2^{n+1}+3\left(n \in \mathbf{N}^{*}\right)$.

Brief solution: Let $\left\{\begin{array}{l}a_{n+1}=5 a_{n}+3 b_{n}+7 \cdots \text { (1) } \\ b_{n+1}=3 a_{n}+5 b_{n} \quad \cdots \text { (2) }\end{array}\right.$
From (1) and (2), we get $1+a_{n+1}+b_{n+1}=8\left(a_{n}+b_{n}+1\right)$, hence $\left\{a_{n}+b_{n}+1\right\}$ forms a geometric sequence with a common ratio of 8.
Thus, $a_{n}+b_{n}+1=8^{n}\left(a_{1}+b_{1}+1\right)=2^{3 n-1}$
(1) - (2) +7 gives $a_{n+1}-b_{n+1}+7=2\left(a_{n}-b_{n}+7\right)$, i.e., $\left\{a_{n}-b_{n}+7\right\}$ forms a geometric sequence with a common ratio of 2.
Hence, $a_{n}-b_{n}+7=2^{n-1}\left(a_{1}-b_{1}+7\right)=2^{n+2} \cdots$
(3) + (4) gives $2 a_{n}+8=2^{3 n-1}+2^{n-2}$, i.e., $a_{n}=2^{3 n-2}+2^{n+1}-4\left(n \in \mathbf{N}^{*}\right)$.
(3) - (4) gives $2 b_{n}-6=2^{3 n-1}-2^{n+2}$, i.e., $b_{n}=2^{3 n-2}-2^{n+1}+3\left(n \in \mathbf{N}^{*}\right)$.