Given the sequence $\left\{a_{n}\right\}$ with the general term
$$
a_{n}=(\sqrt{3}+\sqrt{2})^{2 n}\left(n \in \mathbf{N}_{+}\right) \text {, }
$$

Let $b_{n}=a_{n}+\frac{1}{a_{n}}$.
(1) Find the recurrence relation between $b_{n+2} 、 b_{n+1} 、 b_{n}$;
(2) Find the unit digit of the integer part of $a_{2011}$.
-、(1) From the given, we have
$$
\begin{array}{l}
b_{n}=(\sqrt{3}+\sqrt{2})^{2 n}+(\sqrt{3}-\sqrt{2})^{2 n} \\
\Rightarrow b_{n}=(5+2 \sqrt{6})^{n}+(5-2 \sqrt{6})^{n} .
\end{array}
$$

Then $b_{n+2}=[(5+2 \sqrt{6})+(5-2 \sqrt{6})]$.
$$
\begin{array}{l}
{\left[(5+2 \sqrt{6})^{n+1}+(5-2 \sqrt{6})^{n+1}\right]-} \\
(5+2 \sqrt{6})(5-2 \sqrt{6})\left[(5+2 \sqrt{6})^{n}+(5-2 \sqrt{6})^{n}\right] .
\end{array}
$$

Thus, $b_{n+2}=10 b_{n+1}-b_{n}$.
$$
\begin{array}{l}
\text { (2) Given } b_{1}=10, b_{2}=98, \\
b_{n+2}=10 b_{n+1}-b_{n},
\end{array}
$$

we know that $b_{n}$ is an integer.
If $b_{n}$ is divisible by 10, then $b_{n+2}$ is also divisible by 10.
Since $b_{1}=10$, it follows that $b_{2 k+1}(k \in \mathbf{N})$ is an integer, and the units digit of $b_{2011}$ is 0.

Also, $0<(5-2 \sqrt{6})^{2011}<1$, so the units digit of the integer part of $a_{2011}$ is 9.