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

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

Thus, the original inequality becomes
$$
\begin{array}{l}
x^{2}-1-\frac{\sqrt{2}}{2}(2 \sqrt{2}+\sqrt{2} a) x-\frac{2 \sqrt{2}}{\frac{\sqrt{2}}{2} x}>-3-2 a \\
\Rightarrow(2-x) a>2 x-x^{2}+\frac{4-2 x}{x} \\
=x(2-x)+2 \times \frac{2-x}{x} \\
\Rightarrow a>x+\frac{2}{x} .
\end{array}
$$

The function $f(x)$ is monotonically decreasing on $x \in[1, \sqrt{2}]$. Therefore, the maximum value of $f(x)$ on $x \in[1, \sqrt{2}]$ is $f(1)=3$, which means the range of $a$ is $a>3$.