Given that $a, b, c$ are any complex numbers such that the sum of any two of them is not zero. In the quadratic equations
$$
\begin{array}{l}
x^{2} + a x + b = 0, \\
x^{2} + b x + c = 0, \\
x^{2} + c x + a = 0
\end{array}
$$

any two equations have exactly one root that is the negative of the other.
(1) Find the values of $a^{2} + b^{2} + c^{2}$, $a^{2} b + b^{2} c + c^{2} a$, and $a^{3} b^{2} + b^{3} c^{2} + c^{3} a^{2}$;
(2) Find the values of $a, b, c$ that satisfy the conditions.
(1) Let $x_{1}, x_{2}$ be the roots of equation (1), and $-x_{1}, x_{3}$ be the roots of equation (2), with $x_{2}+x_{3} \neq 0$. Then the two roots of equation (3) have the following four possible cases (all other cases can be reduced to one of these four):
$-x_{1}, -x_{3}, x_{1}, -x_{2}, x_{1}, -x_{1}, -x_{2}, -x_{3}$.
(i) If the roots of equation (3) are $-x_{1}, -x_{3}$, then by
$$
\begin{array}{l}
-x_{1} x_{3}=c, x_{1} x_{3}=a \Rightarrow a+c=0 ; \\
x_{1}+x_{2}=-a, -x_{1}-x_{3}=-c \\
\Rightarrow x_{2}-x_{3}=-a-c=0 \Rightarrow x_{2}=x_{3} .
\end{array}
$$

Thus, equation (1) and equation (3) have two pairs of roots that are opposites of each other, which contradicts the problem statement.
(ii) If the roots of equation (3) are $x_{1}, -x_{2}$, then by
$$
\begin{array}{l}
x_{1} x_{2}=b, -x_{1} x_{2}=a \Rightarrow a+b=0 ; \\
x_{1}+x_{2}=-a, -x_{1}+x_{3}=-b \\
\Rightarrow x_{2}+x_{3}=-a-b=0 .
\end{array}
$$

Thus, equation (1) and equation (2) have two pairs of roots that are opposites of each other, which contradicts the problem statement.
(iii) If the roots of equation (3) are $x_{1}, -x_{1}$, then by
$$
x_{1}-x_{1}=-c \Rightarrow c=0 \Rightarrow x_{1} x_{3}=-c=0 \text {. }
$$

If $x_{1}=0 \Rightarrow a=x_{1} x_{2}=0$.
Then by $x_{1}+x_{2}=-a \Rightarrow x_{2}=0$.
Thus, equation (1) and equation (3) have two pairs of roots that are opposites of each other, which contradicts the problem statement.
(iv) If the roots of equation (3) are $-x_{2}, -x_{3}$, then by
Equation (1) has $x_{1}+x_{2}=-a$;
Equation (2) has $-x_{1}+x_{3}=-b$.
Therefore, $x_{2}+x_{3}=-a-b \neq 0$, meaning equation (1) and equation (2) have only one pair of roots that are opposites of each other, $x_{1}, -x_{1}$.

Similarly, equation (3) and equation (1) have only one pair of roots that are opposites of each other, $x_{2}, -x_{2}$; and equation (2) and equation (3) have only one pair of roots that are opposites of each other, $x_{3}, -x_{3}$.

Thus, when the roots of equation (3) are $-x_{2}, -x_{3}$, the conditions of the problem are satisfied.
$$
\begin{array}{l}
\text { At this point, }\left\{\begin{array}{l}
x_{1}^{2}+a x_{1}+b=0, \\
x_{1}^{2}-b x_{1}+c=0
\end{array} \Rightarrow x_{1}=\frac{c-b}{a+b} .\right. \\
\text { Similarly, } x_{2}=\frac{a-b}{c+a}, x_{3}=\frac{a-c}{b+c} . \\
\text { Since these three equations have six roots, which must be the roots of } \\
\left(x^{2}+a x+b\right)\left(x^{2}+b x+c\right)\left(x^{2}+c x+a\right)=0, \\
x^{6}+(a+b+c) x^{5}+ \\
(a+b+c+b c+c a+a b) x^{4}+ \\
\left(a^{2}+b^{2}+c^{2}+b c+c a+a b+a b c\right) x^{3}+ \\
\left(a^{2} b+b^{2} c+c^{2} a+b c+c a+a b\right) x^{2}+ \\
\left(a b^{2}+b c^{2}+c a^{2}\right) x+a b c=0
\end{array}
$$

That is,
$x^{6}+(a+b+c) x^{5}+$

The sum of the six roots, the sum of the products of every three roots, and the sum of the products of every five roots are all zero. Thus,
$$
\left\{\begin{array}{l}
a+b+c=0, \\
a^{2}+b^{2}+c^{2}+b c+c a+a b+a b c=0, \\
a b^{2}+b c^{2}+c a^{2}=0 .
\end{array}\right.
$$

Therefore, $x_{1} x_{2} x_{3}=\frac{c-b}{a+b} \cdot \frac{b-a}{c+a} \cdot \frac{a-c}{b+c}$
$$
\begin{array}{l}
=\frac{a^{2} b+b^{2} c+c^{2} a-a b^{2}-b c^{2}-c a^{2}}{-a b c} \\
=\frac{a^{2} b+b^{2} c+c^{2} a+a b^{2}+b c^{2}+c a^{2}}{-a b c} \text { (Equation (6) } \\
=\frac{(a+b+c)(b c+c a+a b)-3 a b c}{-a b c} \\
=\frac{-3 a b c}{-a b c}=3 .
\end{array}
$$

By the relationship between the roots and coefficients of the equations, we have
$$
\begin{array}{l}
a b c=-x_{1}^{2} x_{2}^{2} x_{3}^{2}=-9 . \\
(a, b, c) \\
=\left(2 \sqrt{3} \sin 20^{\circ}, 2 \sqrt{3} \sin 40^{\circ},-2 \sqrt{3} \sin 80^{\circ}\right), \\
\left(2 \sqrt{3} \sin 20^{\circ},-2 \sqrt{3} \sin 80^{\circ}, 2 \sqrt{3} \sin 40^{\circ}\right)
\end{array}
$$

and their cyclic permutations.
If $(a, b, c)$
$=\left(2 \sqrt{3} \sin 20^{\circ}, 2 \sqrt{3} \sin 40^{\circ},-2 \sqrt{3} \sin 80^{\circ}\right)$,
then $a^{2}-4 b0, c^{2}-4 a>0$, at this time, equation (1) has a pair of complex conjugate roots, while equations (2) and (3) have real roots, which is a contradiction (since a complex number and a real number cannot be opposites of each other).
Upon verification,
$$
\begin{array}{l}
(a, b, c) \\
=\left(2 \sqrt{3} \sin 20^{\circ},-2 \sqrt{3} \sin 80^{\circ}, 2 \sqrt{3} \sin 40^{\circ}\right)
\end{array}
$$

satisfies the problem statement.
From equations (4), (5), and (7), we get
$b c+c a+a b=(a+b+c)^{2}+a b c=0-9=-9$.
Thus, $a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(b c+c a+a b)$
$=0-2(-9)=18$,
$a^{2} b+b^{2} c+c^{2} a$
$=a^{2} b+b^{2} c+c^{2} a+a b^{2}+b c^{2}+c a^{2}$
$=(a+b+c)(b c+c a+a b)-3 a b c$
$=-3 a b c=-3(-9)=27$,
$a^{3} b^{2}+b^{3} c^{2}+c^{3} a^{2}$
$=\left(a^{2} b+b^{2} c+c^{2} a\right)(b c+c a+a b)-$
$a b c\left(a^{2}+b^{2}+c^{2}+b c+c a+a b\right)$ $=27(-9)-(-9)(18-9)=-162$.
(2) From (1), we have already obtained
$$
\left\{\begin{array}{l}
a+b+c=0, \\
b c+c a+a b=-9, \\
a b c=-9 .
\end{array}\right.
$$

From this, we know that $a, b, c$ are the roots of the equation $y^{3}-9 y+9=0$.
To solve this equation using trigonometric methods, we make the substitution $y=2 \sqrt{3} t$ into $y^{3}-9 y+9=0$ and rearrange to get
$4 t^{3}-3 t+\frac{\sqrt{3}}{2}=0$.
Comparing with the trigonometric identity, we get
$4 \sin ^{3} \alpha-3 \sin \alpha+\sin 3 \alpha=0$.
We just need to take $\sin 3 \alpha=\frac{\sqrt{3}}{2}$, which gives the equation
$4 \sin ^{3} \alpha-3 \sin \alpha+\frac{\sqrt{3}}{2}=0$.
In equation (8), taking $t=\sin \alpha$, we get equation (9).
It is easy to verify that when $\sin \alpha$ is $\sin 20^{\circ}, \sin 40^{\circ}, -\sin 80^{\circ}$, equation (9) always holds.

Thus, $t=\sin 20^{\circ}, \sin 40^{\circ}, -\sin 80^{\circ}$ are the three roots of equation (8).
Since equations (1), (2), and (3) are cyclically symmetric,