(15) (12 points) Inequality
$$
\sin 2 \theta-(2 \sqrt{2}+\sqrt{2} a) \sin \left(\theta+\frac{\pi}{4}\right)-\frac{2 \sqrt{2}}{\cos \left(\theta-\frac{\pi}{4}\right)}>-3-2 a
$$

holds for all $\theta \in\left[0, \frac{\pi}{2}\right]$. Find the range of real number $a$.
15 Let $x=\sin \theta+\cos \theta$, then we have $\sin 2 \theta=x^{2}-1$,
$$
\sin \left(\theta+\frac{\pi}{4}\right)=\cos \left(\theta-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2} x, x \in[1, \sqrt{2}] .
$$

The original inequality is transformed into:
$$
x^{2}-1-(2 \sqrt{2}+\sqrt{2} a) \frac{\sqrt{2}}{2} x-\frac{2 \sqrt{2}}{\frac{\sqrt{2}}{2} x}>-3-2 a \text {. }
$$

That is
$$
x^{2}-1-(2+a) x-\frac{4}{x}+3+2 a>0,
$$

Rearranging gives
$$
(2-x) a>2 x-x^{2}+\frac{4-2 x}{x}=x(2-x)+2 \times \frac{2-x}{x} .
$$

Since $x \in[1, \sqrt{2}], 2-x>0$, we get $a>x+\frac{2}{x}$.
Let $f(x)=x+\frac{2}{x}$, then the function $f(x)$ is monotonically decreasing on $x \in[1, \sqrt{2}]$, so the maximum value of $f(x)$ on $x \in[1, \sqrt{2}]$ is $f(1)=3$. Therefore, the range of $a$ is $a>3$.