3. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n+1}=a_{n}^{2}-2\left(n \in \mathbf{N}_{+}\right) \text {, }
$$

and $a_{1}=a, a_{2012}=b(a 、 b>2)$.
Then $a_{1} a_{2} \cdots a_{2011}=$
(express in terms of $a, b$).
3. $\sqrt{\frac{b^{2}-4}{a^{2}-4}}$.

From $a_{n+1}^{2}-4=a_{n}^{2}\left(a_{n}^{2}-4\right)=\cdots$
$$
=a_{n}^{2} a_{n-1}^{2} \cdots a_{1}^{2}\left(a_{1}^{2}-4\right) \text {, }
$$

then $b^{2}-4=a_{2011}^{2} a_{2010}^{2} \cdots a_{1}^{2}\left(a^{2}-4\right)$.
Therefore, $a_{1} a_{2} \cdots a_{2011}=\sqrt{\frac{b^{2}-4}{a^{2}-4}}$.