1. Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{0}=3, a_{1}=9, a_{n}=4 a_{n-1}-3 a_{n-2}-4 n+2(n \geqslant 2)$. Try to find all non-negative integers $n$, such that $a_{n}$ is divisible by 9.
(Wu Weizhao)
1. It is easy to get $a_{2}=21$. Let $b_{n}=a_{n}-a_{n-1}(n \geqslant 1)$, then $b_{1}=6, b_{2}=12$, and
$$
b_{n}=3 b_{n-1}-4 n+2 \quad(n \geqslant 2) \text {. }
$$

Suppose
$$
b_{n}-\alpha n-\beta=3\left(b_{n-1}-\alpha(n-1)-\beta\right),
$$

where $\alpha, \beta$ are undetermined constants. After rearranging, we get
$$
b_{n}=3 b_{n-1}-2 \alpha n+3 \alpha-2 \beta .
$$

By comparing coefficients, we get $-2 \alpha=-4,3 \alpha-2 \beta=2$, solving these gives $\alpha=\beta=2$, that is
$$
b_{n}-2 n-2=3\left(b_{n-1}-2(n-1)-2\right) .
$$

Thus, the sequence $\left\{b_{n}-2 n-2\right\}$ is a geometric sequence with the first term $b_{1}-4=2$ and common ratio 3, then
$$
b_{n}-2 n-2=2 \times 3^{n-1} \text {, }
$$

which means
$$
b_{n}=2\left(3^{n-1}+n+1\right) \text {. }
$$

Therefore, for $n \geqslant 1$,
$$
\begin{aligned}
a_{n} & =a_{0}+\sum_{i=1}^{n}\left(a_{i}-a_{i-1}\right)=a_{0}+\sum_{i=1}^{n} b_{i} \\
& =3+2 \sum_{i=1}^{n}\left(3^{i-1}+i+1\right) \\
& =3+2 \times \frac{3^{n}-1}{3-1}+n(n+1)+2 n \\
& =3^{n}+n^{2}+3 n+2 .
\end{aligned}
$$

Clearly, $a_{0}=3$ also fits the above formula, hence $a_{n}=3^{n}+n^{2}+3 n+2(n \geqslant 0)$.
Clearly, $9 \nmid a_{0}, 9 \mid a_{1}$. For $n \geqslant 2$, $9 \mid 3^{n}$, so $9 \mid a_{n}$ is equivalent to $9 \mid n^{2}+3 n+2$. After inspection, it is known that $9 \mid n^{2}+3 n+2$ if and only if $n \equiv 7$ or $8(\bmod 9)$.
In summary, all non-negative integers $n$ that make $a_{n}$ divisible by 9 are $1,9 k+7,9 k+8(k=0,1, \cdots)$.