Five. (20 points) In the tetrahedron $P-ABC$, $AB=2$, $PA+PC=4$, $AC=x$, and the maximum volume of $V_{P-ABC}$ is $V$.
(1) Express $V$ in terms of $x$;
(2) If $V_{P-ABC}=\frac{4}{3}$, find the lateral surface area of the tetrahedron $P-ABC$.
(1) As shown in Figure 3, draw $P O \perp$ plane $A B C$ and $P H \perp A C$, with $O$ and $H$ being the feet of the perpendiculars.
$$
\begin{aligned}
& \text { Then } V_{P-A B C} \\
= & \frac{1}{3} S_{\triangle A B C} \cdot P O \\
= & \frac{1}{3} \times \frac{1}{2} A B \cdot A C \sin \angle B A C \cdot P O \\
\leqslant & \frac{1}{3} \times \frac{1}{2} \times 2 x \cdot P O \\
\leqslant & \frac{1}{3} x \cdot P H .
\end{aligned}
$$

The equality in (1) holds if and only if $\angle B A C=90^{\circ}$; the equality in (2) holds if and only if $P H=P O$, i.e., plane $P A C \perp$ plane $A B C$.

In $\triangle P A C$, $P H$ is the altitude to side $A C$, with $A C=x$ and $P A + P C=4$. Using the knowledge of ellipses, we know that when $P A=P C=2$, $P H$ attains its maximum value $\sqrt{2^{2}-\left(\frac{x}{2}\right)^{2}}$.
Therefore, $V_{P-A B C} \leqslant \frac{1}{3} x \sqrt{4-\frac{x^{2}}{4}}$.
The equality holds if and only if $P A=P C=2$, $\angle B A C=90^{\circ}$, and plane $P A C \perp$ plane $A B C$.
Thus, $V=\frac{1}{3} x \sqrt{4-\frac{x^{2}}{4}}=\frac{1}{6} x \sqrt{16-x^{2}}$.
(2) Since $V_{P-A B C} \leqslant V$, then $\frac{4}{3} \leqslant \frac{1}{6} x \sqrt{16-x^{2}}$, which simplifies to $64 \leqslant x^{2}\left(16-x^{2}\right)$.
Therefore, $\left(x^{2}-8\right)^{2} \leqslant 0$.
Solving this, we get $x=2 \sqrt{2}$.
As shown in Figure 4, from the conditions for equality in (1), we have
$$
\begin{array}{l}
P A=P C=2, A C=2 \sqrt{2}, \\
P H=A H=H C=\sqrt{2} .
\end{array}
$$

Thus, $\triangle P A C$ is a right triangle, and $S_{\triangle P A C}=\frac{1}{2} P A \cdot P C=2$.

In the right triangle $\triangle A B C$, since $B C^{2}=A B^{2}+A C^{2}=4+8=12$, we have $B C=2 \sqrt{3}$; since $B H^{2}=A B^{2}+A H^{2}=4+2=6$, we have $B H=\sqrt{6}$.

In the right triangle $\triangle P H B$, since $P B^{2}=P H^{2}+H B^{2}=2+6=8$, we have $P B=2 \sqrt{2}$.

From $P A=A B=2$ and $P B=2 \sqrt{2}$, we know that $\triangle P A B$ is a right triangle. Therefore, $S_{\triangle P A B}=2$.

From $P C=2$, $P B=2 \sqrt{2}$, and $B C=2 \sqrt{3}$, we know that $\triangle P B C$ is a right triangle. Therefore, $S_{\triangle P B C}=2 \sqrt{2}$.
$$
\text { Hence } S_{\text {total}}=S_{\triangle P A B}+S_{\triangle P B C}+S_{\triangle P C A}=4+2 \sqrt{2} \text {. }
$$