1. Given the vector $\overrightarrow{O P}=\left(2 \cos \left(\frac{\pi}{2}+x\right),-1\right)$,
$$
\begin{array}{l}
\overrightarrow{O Q}=\left(-\sin \left(\frac{\pi}{2}-x\right), \cos 2 x\right), \\
f(x)=\overrightarrow{O P} \cdot \overrightarrow{O Q} .
\end{array}
$$

If $a$, $b$, and $c$ are the sides opposite to the acute angles $\angle A$, $\angle B$, and $\angle C$ of $\triangle A B C$, respectively, and satisfy
$$
f(A)=1, b+c=5+3 \sqrt{2}, a=\sqrt{13} \text {, }
$$

then the area $S$ of $\triangle A B C$ is
-1. $\frac{15}{2}$.
From the given conditions, we have
$$
\begin{array}{l}
f(x)=\overrightarrow{O P} \cdot \overrightarrow{O Q} \\
=-2 \cos \left(\frac{\pi}{2}+x\right) \cdot \sin \left(\frac{\pi}{2}-x\right)-\cos 2 x \\
=2 \sin x \cdot \cos x-\cos 2 x=\sqrt{2} \sin \left(2 x-\frac{\pi}{4}\right) \\
\Rightarrow \sqrt{2} \sin \left(2 A-\frac{\pi}{4}\right)=1 \\
\Rightarrow \sin \left(2 A-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}
\end{array}
$$

Since $\angle A$ is an acute angle,
$$
\begin{array}{l}
\Rightarrow-\frac{\pi}{4}<2 \angle A-\frac{\pi}{4}<\frac{3 \pi}{4} \\
\Rightarrow 2 \angle A-\frac{\pi}{4}=\frac{\pi}{4} \Rightarrow \angle A=\frac{\pi}{4} .
\end{array}
$$

Given $b+c=5+3 \sqrt{2}, a=\sqrt{13}$
$$
\begin{array}{l}
\Rightarrow 13=b^{2}+c^{2}-2 b c \cos A \\
\quad=(b+c)^{2}-2 b c-2 b c \cos A \\
\Rightarrow 13=43+30 \sqrt{2}-(2+\sqrt{2}) b c \\
\Rightarrow b c=15 \sqrt{2} .
\end{array}
$$

Therefore, the area of $\triangle A B C$ is
$$
S=\frac{1}{2} b c \sin A=\frac{1}{2} \times 15 \sqrt{2} \times \frac{\sqrt{2}}{2}=\frac{15}{2} .
$$