$$
\begin{array}{l}
\sqrt{2}(2 a+3) \cos \left(\theta-\frac{\pi}{4}\right)+\frac{6}{\sin \theta+\cos \theta}-2 \sin 2 \theta \\
<3 a+6
\end{array}
$$

For $\theta \in\left[0, \frac{\pi}{2}\right]$, the inequality always holds. Find the range of values for $a$.
(Zhang Daocheng, provided)
Let $\sin 0+\cos 0=x$. Then
$$
\begin{aligned}
\cos \left(\theta-\frac{\pi}{4}\right) & =\frac{\sqrt{2}}{2} x, \\
\sin 2 \theta & =x^{2}-1, x \in[1, \sqrt{2}] .
\end{aligned}
$$

Thus, the original inequality can be transformed into
$$
(2 a+3) x+\frac{6}{x}-2\left(x^{2}-1\right)>0.
$$
Rearranging gives
$$
(2 x-3)\left(x+\frac{2}{x}-a\right)>0 \quad(x \in[1, \sqrt{2}]) .
$$

Since $x \in[1, \sqrt{2}]$, we have $2 x-3<0$. Therefore, we need to find the maximum value of $\left(x+\frac{2}{x}\right)$ in the interval $[1, \sqrt{2}]$.
It is easy to see that $f(x)=x+\frac{2}{x}$ is decreasing on $[1, \sqrt{2}]$, so
$$
\left(x+\frac{2}{x}\right)_{\max }=3(x \in[1, \sqrt{2}]) .
$$

Therefore, $a>3$.