Given that $f(x)$ is a function defined on the set of real numbers, and $f(x+2)(1-f(x))=1+f(x)$.

(1) Prove that $f(x)$ is a periodic function.
(2) If $f(1)=2+\sqrt{3}$, find the value of $f(1989)$.
(1) Proof
$$
f(x+2)(1-f(x))=1+f(x)
$$

defined on $R$, we know $f(1) \neq 1$.
$$
\begin{array}{l}
\therefore \quad f(x+2)=\frac{1+f(x)}{1-f(x)}, \\
f(x+4)=-\frac{1}{f(x)}, \\
f(x+8)=f[(x+4)+4]=-\frac{1}{f(x+4)} \\
=f(x) \text {. } \\
\end{array}
$$

Therefore, $f(x)$ is a periodic function with a period of 8.
(2) Solution Since $f(x)$ is a periodic function with a period of 8,

we have,
$$
\begin{aligned}
f(1985) & =2+\sqrt{3} . \\
f(1989) & =f(1985+4)=-\frac{1}{f(1985)} \\
& =-\frac{1}{2+\sqrt{3}}=\sqrt{3}-2 .
\end{aligned}
$$