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

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

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

Since $x \in [1, \sqrt{2}]$, we have $2x - 3 < 0$. Therefore, we need to find the maximum value of $x + \frac{2}{x}$ 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 $\left(x + \frac{2}{x}\right)_{\max} = 3$ for $x \in [1, \sqrt{2}]$.
Thus, $a > 3$.