(12) (13 points) Given a tetrahedron $P-ABC$ with three lateral edges $PA$, $PB$, and $PC$ that are mutually perpendicular, the dihedral angles formed by the lateral faces $PAB$, $PBC$, and $PCA$ with the base $ABC$ have plane angles of $\theta_{1}$, $\theta_{2}$, and $\theta_{3}$, respectively. The area of the base $\triangle ABC$ is $4 \sqrt{3}$.
(1) Prove: $\tan \theta_{1} \cdot \tan \theta_{2} \cdot \tan \theta_{3} \geqslant 2 \sqrt{2}$;
(2) If $\tan \theta_{1}+\tan \theta_{2}+\tan \theta_{3}=3 \sqrt{2}$, find the volume $V_{P-ABC}$ of the tetrahedron.
(12 (1) Let's assume $P A=a, P B=b, P C=c$, construct $P N \perp A B$ at $N$, and connect $C N$. Then it is easy to know $\angle P N C=\theta_{1}$,
$$
\begin{aligned}
\tan \theta_{1} & =\tan \angle P N C=\frac{P C}{P N} \\
& =\frac{c}{\frac{a b}{\sqrt{a^{2}+b^{2}}}} \\
& =\frac{c}{a b} \cdot \sqrt{a^{2}+b^{2}} .
\end{aligned}
$$

Similarly, $\tan \theta_{2}=\frac{a}{b c} \sqrt{b^{2}+c^{2}}, \tan \theta_{3}=\frac{b}{a c} \sqrt{c^{2}+a^{2}}$.
Therefore, $\tan \theta_{1} \cdot \tan \theta_{2} \cdot \tan \theta_{3}$
$$
\begin{array}{l}
=\frac{1}{a b c} \sqrt{a^{2}+b^{2}} \cdot \sqrt{b^{2}+c^{2}} \cdot \sqrt{c^{2}+a^{2}} \\
\geqslant \frac{1}{a b c} \sqrt{2 a b} \cdot \sqrt{2 b c} \cdot \sqrt{2 c a}=2 \sqrt{2},
\end{array}
$$

with equality holding if and only if $a=b=c$.
(2)
$$
\begin{aligned}
3 \sqrt{2} & =\tan \theta_{1}+\tan \theta_{2}+\tan \theta_{3} \\
& \geqslant 3 \sqrt[3]{\tan \theta_{1} \cdot \tan \theta_{2} \cdot \tan \theta_{3}} \\
& \geqslant 3 \sqrt{2},
\end{aligned}
$$

with equality holding if and only if $\tan \theta_{1}=\tan \theta_{2}=\tan \theta_{3}$, i.e., $a=b=c$. From $a=b=c$ and $S_{\triangle A B C}=4 \sqrt{3}$, we get $\frac{\sqrt{3}}{4} \times(\sqrt{2} a)^{2}=4 \sqrt{3}$, thus $a=$ $b=c=2 \sqrt{2}$, hence
$$
V_{P-A B C}=V_{C-P A B}=\frac{1}{6} a b c=\frac{8 \sqrt{2}}{3} .
$$