7. Given $x_{1}=1, x_{2}=2$, and $x_{n+12}=\frac{x_{n+1}^{2}-7}{x_{n}}\left(n \in \mathbf{N}^{*}\right)$, then the smallest positive period of the sequence $\left\{x_{n}\right\}$ is . $\qquad$
7. 3 From $x_{1}=1, x_{2}=2$, we get
$$
\begin{array}{l}
x_{3}=\frac{2^{2}-7}{1}=-3, \\
x_{1}=\frac{(-3)^{2}-7}{2}=1, \\
x_{:}=\frac{1^{2}-7}{-3}=2, \\
\ldots \ldots
\end{array}
$$

It can be seen that $\left\{x_{n}\right\}$ is a periodic sequence with a period of 3.