11. Find the range of real number $a$ such that
$$
\sin 2 \theta-2 \sqrt{2} a \cos \left(\theta-\frac{\pi}{4}\right)-\frac{\sqrt{2} a}{\sin \left(\theta+\frac{\pi}{4}\right)}>-3-a^{2}
$$

holds for $\theta \in\left[0, \frac{\pi}{2}\right]$.
II. 11. Let $x=\sin \theta+\cos \theta\left(\theta \in\left[0, \frac{\pi}{2}\right]\right)$.
Then $x \in[1, \sqrt{2}], \sin 2 \theta=x^{2}-1$,
$$
\sin \left(\theta+\frac{\pi}{4}\right)=\cos \left(\theta-\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2} x \text {. }
$$

Therefore, the original inequality can be transformed into
$$
x^{2}-2 a x-\frac{2 a}{x}+2+a^{2}>0 \text {, }
$$

which is equivalent to $(a-x)\left(a-\left(x+\frac{2}{x}\right)\right)>0$.
Let $f(x)=x+\frac{2}{x}$, we know that $f(x)$ is monotonically decreasing in the interval $[1, \sqrt{2}]$.
Thus, $f(x)_{\max }=f(1)=3$.
If $1 \leqslant a \leqslant 3$, then when $x=1$, equation (1) does not hold, which does not meet the problem's conditions;

If $a>3$, then for all $x \in[1, \sqrt{2}]$, $a-x$ and $a-\left(x+\frac{2}{x}\right)$ have the same sign, equation (1) always holds, which meets the problem's requirements.
Therefore, $a \in(-\infty, 1) \cup(3,+\infty)$.