8. In the sequence $\left\{a_{n}\right\}$, $a_{1}=1, a_{2}=2$, and $a_{n+2}=7 a_{n-1}-12 a_{n}$, then $a_{n}=$
8. $a_{n}=2 \cdot 3^{n-1}-4^{n-1}\left(n \in \mathbf{N}^{*}\right)$.
Brief solution: The characteristic equation of the given recurrence relation is $x^{2}=7 x-12$, with its two roots being 3 and 4. Therefore, the general term of the sequence $\left\{a_{n}\right\}$ can be assumed as $a_{n}=A \cdot 3^{n}+B \cdot 4^{n}$, where $a_{1}=1, a_{2}=2$. Thus, from $\left\{\begin{array}{l}3 A+4 B=1 \\ 9 A+16 B=2\end{array}\right.$ we get $\left\{\begin{array}{l}A=\frac{2}{3} \\ B=-\frac{1}{4}\end{array}\right.$, so $a_{n}=2 \cdot 3^{n-1}-4^{n-1}\left(n \in \mathbf{N}^{*}\right)$.