12. Given the sequence $\left\{a_{n}\right\}$, where $a_{1}=1, a_{2}=2, a_{n+1}=3 a_{n}-2 a_{n-1}, a_{2002}=$
12. $a_{2002}=2^{2001}$.

From $a_{n+1}=3 a_{n}-2 a_{n-1}$, we can respectively get $a_{n+1}-2 a_{n}=a_{n}-2 a_{n-1}, a_{n}-2 a_{n-1}=0$, so $\left\{a_{n}\right\}$ is a geometric sequence with the first term $a_{1}$ and common ratio 2. Therefore, $a_{n}=2^{n-1}$.