1. Let $f(x)$ be a function defined on $\mathbf{R}$, such that for any real number $x$, $f(x+3) \cdot f(x-4)=-1$. Also, when $0 \leqslant x<7$, $f(x)=\log _{2}(9-x)$, then the value of $f(-100)$ is $\qquad$ .
$$
f(x+3)=-\frac{1}{f(x-4)} \Rightarrow f(x+7)=-\frac{1}{f(x)} \Rightarrow f(x+14)=f(x) \text {, }
$$

Then $f(x)$ is a periodic function with period $T=14$.
So $f(-100)=f(-100+14 \times 8)=f(12)=-\frac{1}{f(5)}=-\frac{1}{\log _{2} 4}=-\frac{1}{2}$.