1) Given the vectors $\overrightarrow{O P}=\left(2 \cos \left(\frac{\pi}{2}+x\right),-1\right), \overrightarrow{O Q}=\left(-\sin \left(\frac{\pi}{2}-\right.\right.$ $x), \cos 2 x), f(x)=\overrightarrow{O P} \cdot \overrightarrow{O Q}$. If $a, b, c$ are the sides opposite to the angles $A, B, C$ of the acute triangle $\triangle A B C$, and it satisfies $f(A)=1, b+c=5+3 \sqrt{2}, a=\sqrt{13}$, then the area $S$ of $\triangle A B C$ is $S=$ . $\qquad$
(1) $\frac{15}{2}$ Hint: From the given conditions,
$$
\begin{aligned}
f(x) & =\overrightarrow{O P} \cdot \overrightarrow{O Q} \\
& =-2 \cos \left(\frac{\pi}{2}+x\right) \sin \left(\frac{\pi}{2}-x\right)-\cos 2 x \\
& =2 \sin x \cos x-\cos 2 x \\
& =\sqrt{2} \sin \left(2 x-\frac{\pi}{4}\right),
\end{aligned}
$$

So $\sqrt{2} \sin \left(2 A-\frac{\pi}{4}\right)=1, \sin \left(2 A-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}$.
Since $A$ is an acute angle, $-\frac{\pi}{4}<2 A-\frac{\pi}{4}<\frac{3 \pi}{4}$, thus $2 A-\frac{\pi}{4}=$ $\frac{\pi}{4}, A=\frac{\pi}{4}$.
Given $b+c=5+3 \sqrt{2}, a=\sqrt{13}$, we have
$$
13=b^{2}+c^{2}-2 b c \cos A=(b+c)^{2}-2 b c-2 b c \cos A,
$$

which means
$$
13=43+30 \sqrt{2}-(2+\sqrt{2}) b c .
$$

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