(2) Let the sequence $\left\{a_{n}\right\}$ satisfy: $a_{1}=\frac{1}{2}, a_{n+1}=\frac{1+a_{n}}{1-a_{n}} \quad(n \geqslant 1)$, then $a_{2008}=$
(2) According to the recursive relation, we can calculate that $a_{2}=3, a_{3}=-2, a_{4}=-\frac{1}{3}, a_{5}=\frac{1}{2}=a_{1}$, and it can be proven that: when $n \geqslant 1$, $a_{n+4}=a_{n}$. Therefore, the sequence $\left\{a_{n}\right\}$ is a periodic sequence with a period of 4, hence $a_{2008}=a_{4}=-\frac{1}{3}$.