$3 \cdot 30$ (1) Let $n$ be a prime number greater than 3. Find the value of $\left(1+2 \cos \frac{2 \pi}{n}\right)$ $\left(1+2 \cos \frac{4 \pi}{n}\right)\left(1+2 \cos \frac{6 \pi}{n}\right) \cdots \cdot\left(1+2 \cos \frac{2 n \pi}{n}\right)$;
(2) Let $n$ be a natural number greater than 3, find the value of $\left(1+2 \cos \frac{\pi}{n}\right)\left(1+2 \cos \frac{2 \pi}{n}\right)\left(1+2 \cos \frac{3 \pi}{n}\right) \cdots$. $\left(1+2 \cos \frac{(n-1) \pi}{n}\right)$.
[Solution] (1) Let $w=e^{\frac{2 \pi i}{n}}$. Clearly, $w^{n}=1, w^{-\frac{n}{2}}=-1, w^{k}+w^{-k}=2 \cos \frac{2 k \pi}{n}$. Therefore, we have
$$
\begin{aligned}
& \prod_{k=1}^{n}\left(1+2 \cos \frac{2 k \pi}{n}\right)=\prod_{k=1}^{n}\left(1+w^{k}+w^{-k}\right) \\
= & \prod_{k=1}^{n} w^{-k}\left(w^{k}+w^{2 k}+1\right) \\
= & w^{-\frac{n(n+1)}{2}} \prod_{k=1}^{n-1} \frac{1-w^{3 k}}{1-w^{k}} \cdot\left(w^{n}+w^{2 n}+1\right) \\
= & (-1)^{n+1} \cdot 3 \cdot \prod_{k=1}^{n-1} \frac{1-w^{3 k}}{1-w^{k}}
\end{aligned}
$$

Since $n$ is a prime number greater than 3, $(-1)^{n+1}=1$, and
$$
3 \times 1,3 \times 2, \cdots, 3(n-1) \quad(\bmod n)
$$

covers all the residue classes of $n$, thus we have
$$
\prod_{k=1}^{n-1}\left(1-w^{3 k}\right)=\prod_{k=1}^{n-1}\left(1-w^{k}\right).
$$

Therefore,
$$
\prod_{k=1}^{n}\left(1+2 \cos \frac{2 k \pi}{n}\right)=3.
$$
(2) Since the $2n$ roots of $z^{2 n}-1=0$ are $\pm 1$ and $z_{k}=e^{\pm \frac{k \pi i}{n}}(k=1,2, \cdots, n-1)$, we have
$$
\begin{aligned}
z^{2 n}-1 & =\left(z^{2}-1\right) \prod_{k=1}^{n-1}\left(z-e^{\frac{k \pi i}{n}}\right)\left(z-e^{-\frac{k \pi i}{n}}\right) \\
& =\left(z^{2}-1\right) \prod_{k=1}^{n-1}\left(z^{2}+1-2 z \cos \frac{k \pi}{n}\right)
\end{aligned}
$$

Let $z=e^{\frac{2 \pi i}{3}}$, then $z^{2}+1=-z$, so we have
$$
\begin{array}{l}
z^{2 n}-1=\left(z^{2}-1\right)(-z)^{n-1} \prod_{k=1}^{n-1}\left(1+2 \cos \frac{k \pi}{n}\right), \\
\prod_{k=1}^{n-1}\left(1+2 \cos \frac{k \pi}{n}\right)=\frac{z^{2 n}-1}{\left(z^{2}-1\right)(-z)^{n-1}}
\end{array}
$$

When $n=3 k$, $z^{2 n}=z^{6 k}=\left(e^{\frac{2 \pi i}{3}}\right)^{6 k}=1$, thus
$$
\prod_{k=1}^{n-1}\left(1+2 \cos \frac{k \pi}{n}\right)=0.
$$

When $n=3 k+1$, $z^{2 n}=z^{6 k+2}=z^{2}$, thus
$$
\begin{aligned}
\prod_{k=1}^{n-1}\left(1+2 \cos \frac{k \pi}{n}\right) & =\frac{z^{2}-1}{\left(z^{2}-1\right)(-z)^{3 k}}=(-1)^{3 k} \\
& =(-1)^{n-1}.
\end{aligned}
$$

When $n=3 k+2$, $z^{2 n}=z^{6 n+4}=z$, thus
$$
\begin{aligned}
\prod_{k=1}^{n-1}\left(1+2 \cos \frac{k \pi}{n}\right) & =\frac{z-1}{\left(z^{2}-1\right)(-z)^{3 k+1}} \\
& =(-1)^{3 k+1} \cdot \frac{z-1}{\left(z^{2}-1\right) \cdot z} \\
& =(-1)^{3 k+1} \frac{z-1}{z^{3}-z}=(-1)^{3 k+1} \frac{z-1}{1-z} \\
& =(-1)^{3 k+2}=(-1)^{n}.
\end{aligned}
$$

In summary, we have
$$
\prod_{k=1}^{n-1}\left(1+2 \cos \frac{k \pi}{n}\right)=\left\{\begin{array}{l}
0, \text{ when } n=3 k, \\
(-1)^{n-1}, \text{ when } n=3 k+1, k \in \mathbb{N}, \\
(-1)^{n}, \text{ when } n=3 k+2.
\end{array}\right.
$$