1. Given the parabola equation $y=-\frac{x^{2}}{2}+h(h>0)$, and point $P(0, h)$, and the parabola intersects at points $A, B$. The inclination angles of lines $PA$ and $PB$ are complementary. Then the maximum area of $\triangle PAD$ is

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$$
=1 . \frac{64 \sqrt{3}}{9} \text {. }
$$

Given $P(2,4)$ lies on $y=-\frac{x^{2}}{2}+h$, we get $h=6$.
The equations of lines $P A$ and $P B$ are $y-4=-k(x-2)$ and $y-4=k(x-2)$, respectively.

Substituting $y-4=k(x-2)$ into $y=-\frac{x^{2}}{2}+6$, we get $x^{2}+2kx-4k-1=0$.
Thus, $x_{1}=2k-2, x=-2k-2$,
and $y_{B}=k(x_{B}-2)+4, y_{A}=-k(x_{A}-2)+4$.
From this, we get $k_{A B}=\frac{y_{B}-y_{A}}{x_{B}-x_{A}}=2$.
Assume the equation of line $A B$ is $y=2x+m (m>0)$. Substituting into $y=-\frac{x^{2}}{2}+6$ gives
$$
x^{2}+4x+2m-12=0, \Delta=16-4(2m-12)>0,
$$
$0<m<8$.
$$
|A B|^{2}=5\left(x_{A}-x_{B}\right)^{2}=40(8-m) \text {, }
$$

The distance from point $P(2,4)$ to line $A B$ is
$$
\begin{array}{l}
d=\frac{|2 \times 2-4+m|}{\sqrt{5}}=\frac{|m|}{\sqrt{5}}=\frac{m}{\sqrt{5}} . \\
\begin{array}{l} 
S_{\triangle P A B}^{2}=\frac{1}{4} \cdot\left(\frac{m}{\sqrt{5}}\right)^{2} \cdot 40(8-m) \\
=8 \cdot \frac{m}{2} \cdot \frac{m}{2} \cdot(8-m) \\
\leqslant 8 \cdot\left(\frac{8}{3}\right)^{3}(\because 0<m<8),
\end{array} \\
\therefore S_{\triangle P A B} \leqslant \frac{64 \sqrt{3}}{9} .
\end{array}
$$