II. (20 points) The sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{1}=2, a_{n}=\frac{a_{n-1}^{2}}{a_{n-2}}(n=3,4, \cdots) \text {. }
$$

Suppose $a_{2} 、 a_{5}$ are positive integers, and $a_{5} \leqslant 2010$. Find all possible values of $a_{5}$.
From the given, $\frac{a_{n}}{a_{n-1}}=\frac{a_{n-1}}{a_{n-2}}$. Therefore, the sequence $\left\{a_{n}\right\}$ is a geometric sequence.
Let $a_{2}=x$. Then the common ratio of the sequence $\left\{a_{n}\right\}$ is $\frac{x}{2}$.
Thus, $a_{5}=2\left(\frac{x}{2}\right)^{4}=\frac{x^{4}}{8}$.
Since $a_{5}$ is a positive integer, then, $21 x$.
Let $x=2 y$. Then $a_{5}=2 y^{4}$.
Therefore, $2 y^{4} \leqslant 2010$, which means $y \leqslant 5$.
Hence, $y=1,2,3,4,5$.
Thus, $a_{5}=2,32,162,512,1250$.