12. As shown in the figure, $\triangle A B C$ is an isosceles triangle with the vertex angle $\frac{\pi}{5}$. This triangle is denoted as $\triangle_{1}$, and its vertex angle is denoted as $\alpha_{1}\left(\alpha_{1}=\angle B A C\right)$. Using one of the base angles of $\triangle_{1}$ $(\angle A B C)$ as the vertex angle, construct an isosceles triangle $\triangle_{2}$ (i.e., $\triangle B C D$), with its vertex angle denoted as $\alpha_{2}\left(\alpha_{2}=\angle D B C\right)$; then, using one of the base angles of $\triangle_{2}$ $(\angle B C D)$ as the vertex angle, construct an isosceles triangle $\triangle_{3}$ (i.e., $\triangle C E B$), with its vertex angle denoted as $\alpha_{3}\left(\alpha_{3}=\angle B C E\right)$. Continue this process to construct isosceles triangles $\triangle_{1}, \triangle_{2}, \cdots, \triangle_{n}, \cdots$, and the corresponding vertex angles form the sequence $\left\{a_{n}\right\}$.
(1) Find $\alpha_{2}, \alpha_{3}, \alpha_{4}$;
(2) Find the general term formula for $\alpha_{n}$.
12. (1) According to the problem, given $\alpha_{1}=\frac{\pi}{5}$, we have $\alpha_{2}=\frac{1}{2}\left(\pi-\alpha_{1}\right)=\frac{1}{2}\left(\pi-\frac{\pi}{5}\right)=\frac{2 \pi}{5}$,
$$
\alpha_{3}=\frac{1}{2}\left(\pi-\alpha_{2}\right)=\frac{1}{2}\left(\pi-\frac{2 \pi}{5}\right)=\frac{3 \pi}{10}, \alpha_{4}=\frac{1}{2}\left(\pi-\alpha_{3}\right)=\frac{1}{2}\left(\pi-\frac{3 \pi}{10}\right)=\frac{7 \pi}{20} \text {. }
$$
(2) By the sum of the interior angles of a triangle, we have $\alpha_{n}=\frac{1}{2}\left(\pi-\alpha_{n-1}\right), \alpha_{n+1}=\frac{1}{2}\left(\pi-\alpha_{n}\right)$.
Subtracting these, we get $a_{n+1}-a_{n}=-\frac{1}{2}\left(\alpha_{n}-a_{n-1}\right)$.
This indicates that the sequence $\left\{\alpha_{n+1}-a_{n}\right\}$ is a geometric sequence with the first term $\alpha_{2}-\alpha_{1}=\frac{\pi}{5}$ and common ratio $q=-\frac{1}{2}$, so $\alpha_{n+1}-\alpha_{n}=\left(\alpha_{2}-\right.$ $\left.\alpha_{1}\right)\left(-\frac{1}{2}\right)^{n-1}=\left(-\frac{1}{2}\right)^{n-1} \cdot \frac{\pi}{5}$
Thus, $\alpha_{n}=\left(\alpha_{n}-\alpha_{n-1}\right)+\cdots+\left(\alpha_{2}-\alpha_{1}\right)+\alpha_{1}=\left[\left(-\frac{1}{2}\right)^{n-2}+\cdots+\left(-\frac{1}{2}\right)+1\right] \frac{\pi}{5}+\frac{\pi}{5}$ $=\frac{\pi}{3}+\left(-\frac{1}{2}\right)^{n} \cdot \frac{4 \pi}{15}$.
This is the general term of the sequence.