(8) The function $f(x)$ satisfies $f(x+3)=-\frac{1}{f(x)}$ for any real number $x$, and $f(1)=\frac{1}{2}$, then $f(2008)=$ $\qquad$
8 Since
$$
f(x+6)=-\frac{1}{f(x+3)}=f(x),
$$

thus $f(x)$ is a periodic function with a period of 6. Therefore,
$$
f(2008)=f(6 \times 334+4)=f(4)=-\frac{1}{f(1)}=-2 .
$$