Mi, (Full marks 20 points) Find the range of real values of $a$ such that the inequality
$$
\begin{array}{l}
\sin 2 \theta-(2 \sqrt{2}+a \sqrt{2}) \sin \left(\theta+\frac{\pi}{4}\right) \\
-\frac{2 \sqrt{2}}{\cos \left(\theta-\frac{\pi}{4}\right)}>-3-2 a
\end{array}
$$

holds for all $\theta \in\left[0, \frac{\pi}{2}\right]$.
Three, let $x=\sin \theta+\cos 9$. From $\theta \leq\left[0, \frac{\pi}{2}\right]$, we get $x \in[1, \sqrt{2}]$. Thus, $\sin 2 \theta=x^{2}-1, \sin \left(\theta+\frac{\pi}{4}\right)=$ $\cos \left(\theta-\frac{\pi}{4}\right)=\frac{x}{\sqrt{2}}$. The original inequality becomes
$$
x^{2}-1-(2+a) x-\frac{4}{x}+3+2 a>0 \text {, }
$$

which simplifies to $(x-2)\left(x+\frac{2}{x}-a\right)>0$.
$$
\because 1 \leqslant x \leqslant \sqrt{2}, \therefore x-2x+\frac{2}{x}, x \in[1, \sqrt{2}]$.
To ensure the original inequality always holds, we need $a>\left(x+\frac{2}{x}\right)_{\operatorname{max}}$. It is easy to see that the function $f(x)=x+\frac{2}{x}$ is decreasing on $[1, \sqrt{2}]$.
Therefore, $f(x)_{\max }=f(1)=3$. Hence, $a>3$.