3.2.8* The sequence $\left\{a_{n}\right\}$ satisfies: $a_{1}=1, a_{2}=2, a_{n+2}+a_{n+1}+a_{n}=0$. Find the general term formula $a_{n}$.
The characteristic equation is $x^{2}+x+1=0$. The characteristic roots are solved as $x_{1}=\cos \frac{2 \pi}{3}+\mathrm{i} \sin \frac{2 \pi}{3}, x_{2}=$ $\cos \frac{2 \pi}{3}-\mathrm{i} \sin \frac{2 \pi}{3}$. Therefore, $a_{n}=A_{1}\left(\cos \frac{2 n \pi}{3}+i \sin \frac{2 n \pi}{3}\right)+A_{2}\left(\cos \frac{2 n \pi}{3}-\mathrm{i} \sin \frac{2 n \pi}{3}\right)$. Substituting the initial conditions $a_{1}=1, a_{2}=2$, we get
$$
\left\{\begin{array}{l}
\Lambda_{1}\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)+\Lambda_{2}\left(-\frac{1}{2}-\frac{\sqrt{3}}{2} \mathrm{i}\right)=1, \\
\Lambda_{1}\left(-\frac{1}{2}-\frac{\sqrt{3}}{2} \mathrm{i}\right)+\Lambda_{2}\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)=2 .
\end{array}\right.
$$

Solving, we get $A_{1}=\frac{9+\sqrt{3} \mathrm{i}}{6}, A_{2}=\frac{-9-\sqrt{3} \mathrm{i}}{6}$. Therefore,
$$
\begin{aligned}
a_{n} & =\frac{-9+\sqrt{3} i}{6}\left(\cos \frac{2 n \pi}{3}+\mathrm{i} \sin \frac{2 n \pi}{3}\right)+\frac{-9-\sqrt{3} i}{6}\left(\cos \frac{2 n \pi}{3}-\mathrm{i} \sin \frac{2 n \pi}{3}\right) \\
& =-3 \cos \frac{2 n \pi}{3}-\frac{\sqrt{3}}{3} \sin \frac{2 n \pi}{3} .
\end{aligned}
$$