(1) Let real numbers $x_{1}, x_{2}, \cdots, x_{1997}$ satisfy the following two conditions:
(1) $-\frac{1}{\sqrt{3}} \leqslant x_{i} \leqslant \sqrt{3}(i=1,2, \cdots, 1997)$;
(2) $x_{1}+x_{2}+\cdots+x_{1977}=-318 \sqrt{3}$.

Find: $x_{1}^{12}+x_{2}^{12}+\cdots+x_{1997}^{12}$'s maximum value, and explain the reason.
Solve for any set of $x_{1}, x_{2}, \cdots, x_{1997}$ that satisfies the given conditions, if there exist such $x_{i}$ and $x_{j}$ that
$$
\sqrt{3}>x_{i} \geqslant x_{j}>-\frac{1}{\sqrt{3}}
$$

then we denote
$$
\begin{array}{c}
m=\frac{1}{2}\left(x_{i}+x_{j}\right) \\
h_{0}=\frac{1}{2}\left(x_{i}-x_{j}\right)=x_{i}-m=m-x_{j}
\end{array}
$$

We observe that
$$
(m+h)^{12}+(m-h)^{12}=2 \sum_{0 \leq 6=2 l_{\leq} 12} \mathrm{C}_{12}^{k} \cdot m^{12-k} \cdot h^{k}
$$

increases as $h>0$ increases. We agree to take
$$
h=\min \left\{\sqrt{3}-m, m-\left(-\frac{1}{\sqrt{3}}\right)\right\}
$$

and replace $x_{i}$ and $x_{j}$ with $x^{\prime}{ }_{i}=m+h, x^{\prime}{ }_{j}=m-h$. The sum of the variables remains unchanged, while the sum of the 12th powers of the variables increases.

Therefore, the maximum value of the sum of the 12th powers can only be achieved in the following scenario: at most one variable takes a value in $\left(-\frac{1}{\sqrt{3}}, \sqrt{3}\right)$, and the rest of the variables take values of $-\frac{1}{\sqrt{3}}$ or $\sqrt{3}$. Suppose there are $u$ variables taking the value $-\frac{1}{\sqrt{3}}$, $v$ variables taking the value $\sqrt{3}$, and $w(=0$ or 1$)$ variables taking a value in $\left(-\frac{1}{\sqrt{3}}, \sqrt{3}\right)$ (if there is one, denote this value as $t$).
Thus,
$$
\left\{\begin{array}{l}
u+v+w=1997 \\
-\frac{1}{\sqrt{3}} u+\sqrt{3} v+t w=-318 \sqrt{3}
\end{array}\right.
$$

From this, we get
$$
4 v+(\sqrt{3} t+1) w=1043
$$

Since $(\sqrt{3} t+1) w=1043-4 v$ is an integer, and
$$
0 \leqslant(\sqrt{3} t+1) w<4
$$

Therefore, $(\sqrt{3} t+1) w$ is the remainder when 1043 is divided by 4. According to this, we find
$$
v=260, t=\frac{2}{\sqrt{3}}, u=1736
$$

According to the above discussion, the maximum value of $x_{1}^{12}+x_{2}^{12}+\cdots+x_{1997}^{12}$ is
$$
\begin{array}{l}
\left(-\frac{1}{\sqrt{3}}\right)^{12} u+(\sqrt{3})^{12} v+t^{12}= \\
\frac{1736+4096}{729}+729 \times 260= \\
8+189540=189548
\end{array}
$$