1. Given the function $f(x)$ satisfies $f(1)=2$, and
$$
f(x+1)=\frac{1+f(x)}{1-f(x)}
$$

for any $x$ in its domain. Then $f(2016)=$
$\qquad$
$-1 . \frac{1}{3}$.
From $f(x+1)=\frac{1+f(x)}{1-f(x)}$, it is easy to get
$$
f(x+2)=-\frac{1}{f(x)}, f(x+4)=f(x) \text {. }
$$

Thus, $f(x)$ is a periodic function with a period of 4.
Given $f(2)=-3, f(3)=-\frac{1}{2}, f(4)=\frac{1}{3}$
$$
\Rightarrow f(2016)=f(4)=\frac{1}{3} \text {. }
$$