(1) (Full marks 20 points) Given the equation about $x$
$$
\left(m^{2}-1\right) x^{2}-3(3 m-1) x+18=0
$$

has two positive integer roots ($m$ is an integer). The three sides $a$, $b$, $c$ of $\triangle A B C$ satisfy $c=2 \sqrt{3}$, $m^{2}+a^{2} m-8 a=0$, $m^{2}+b^{2} m-8 b=0$. Find:
(1) the value of $m$;
(2) the area of $\triangle A B C$.
(1) The equation has two real roots, then $m^{2}-1 \neq 0$. Solving the equation gives $x_{1}=\frac{6}{m+1}, x_{2}=\frac{3}{m-1}$.
From the problem, we have
$\left\{\begin{array}{l}m+1=1,2,3,6, \\ m-1=1,3,\end{array}\right.$
which means $\left\{\begin{array}{l}m=0,1,2,5, \\ m=2,4 .\end{array}\right.$
Therefore, $m=2$.
(2) Substituting $m=2$ into the two equations, we get
$$
a^{2}-4 a+2=0, b^{2}-4 b+2=0 \text {. }
$$

When $a=b$, $a=b=2 \pm \sqrt{2}$.
When $a \neq b$, $a, b$ are the two roots of the equation $x^{2}-4 x+2=0$, and $\Delta>0$, by Vieta's formulas we get $a+b=4>0, a b=2>0$, so $a>0, b>0$.
(i) When $a=b=2+\sqrt{2}, c=2 \sqrt{3}$, since $2(2+\sqrt{2})>$ $2 \sqrt{3}$, a triangle can be formed.
$$
\begin{array}{l}
=\sqrt{y+12 \sqrt{2}} \text {. } \\
\end{array}
$$
(ii) When $a=b=2-\sqrt{2}, c=2 \sqrt{3}$, since $2(2-\sqrt{2})<$ $2 \sqrt{3}$, a triangle cannot be formed, which is not suitable, so it is discarded.
(iii) When $a \neq b, c=2 \sqrt{3}$, since $a^{2}+b^{2}=(a+b)^{2}$ $-2 a b=16-4=12=c^{2}$, $\triangle A B C$ is a right triangle, H. $\angle C=90^{\circ}, S_{\triangle A B C}=\frac{1}{2} a b=1$.

In summary (i), (ii), and (iii), the area of $\triangle A B C$ is 1 or $\sqrt{9+12 \sqrt{2}}$.