1. Given the function $f(x)$ satisfies: $f(1)=2, f(x+1)=\frac{1+f(x)}{1-f(x)}$ for any $x$ in the domain, then $f(2016)=$
$$
\begin{array}{l}
f(x+2)=\frac{1+f(x+1)}{1-f(x+1)}=\frac{1+\frac{1+f(x)}{1-f(x)}}{1-\frac{1+f(x)}{1-f(x)}}=-\frac{1}{f(x)} \Rightarrow f(x+4)=-\frac{1}{f(x+2)}=f(x) \\
\Rightarrow f(2016)=f(0). \text{ From } f(1)=\frac{1+f(0)}{1-f(0)}=2 \Rightarrow f(0)=\frac{1}{3}, \text{ so } f(2016)=\frac{1}{3}.
\end{array}
$$