1. When $n$ is a positive integer, the function $f$ satisfies:
$$
\begin{array}{l}
f(n+3)=\frac{f(n)-1}{f(n)+1}, \\
f(1) \neq 0 \text { and } f(1) \neq \pm 1 .
\end{array}
$$

Then the value of $f(11) f(2021)$ is $\qquad$
$$
\text { - 1. }-1 \text {. }
$$

From the given information, we have
$$
\begin{array}{l}
f(n+6)=-\frac{1}{f(n)} . \\
\text { Therefore, } f(11) f(2021)=f(11) f(5)=-1 .
\end{array}
$$