8. Given that $f(x)$ is a function defined on $\mathbf{R}$, and satisfies $f(x+2)[1-f(x)]=1+f(x), f(1)=$ 9997, then the value of $f(2009)$ is $\qquad$ .
8. 9997. From the white variable values 2009 and 1, it is easy to infer that the function $f(x)$ is a periodic function. Given that $f(x) \neq 1$, we have $f(x+2)=\frac{1+f(x)}{1-f(x)}$.
Therefore, $f(x+4)=\frac{1+f(x+2)}{1-f(x+2)}=\frac{1+\frac{1+f(x)}{1-f(x)}}{1-\frac{1+f(x)}{1-f(x)}}=-\frac{1}{f(x)}$. That is, $f(x+4)=-\frac{1}{f(x)}, f(x+8)=-\frac{1}{f(x+4)}$, so we know $f(x+8)=f(x)$, meaning the function $y=f(x)$ is a periodic function with a period of 8. Thus, $f(2009)=f(1+8 \times 251)=f(1)=9997$.