9.68 Find the range of real number $a$ such that for any real number $x$ and any $\theta \in \left[0, \frac{\pi}{2}\right]$, the following inequality always holds:
$$(x+3+2 \sin \theta \cos \theta)^{2}+(x+a \sin \theta+a \cos \theta)^{2} \geqslant \frac{1}{8}$$
[Solution] It is easy to know that the original inequality is equivalent to
$$(3+2 \sin \theta \cos \theta-a \sin \theta-a \cos \theta)^{2} \geqslant \frac{1}{4}$$

for any $\theta \in\left[0, \frac{\pi}{2}\right]$.
From (1), we get
$$a \geqslant \frac{3+2 \sin \theta \cos \theta+\frac{1}{2}}{\sin \theta+\cos \theta}$$, for any $\theta \in\left[0, \frac{\pi}{2}\right]$
or $$a \leqslant \frac{3+2 \sin \theta \cos \theta-\frac{1}{2}}{\sin \theta+\cos \theta}$$, for any $\theta \in\left[0, \frac{\pi}{2}\right]$,
Under condition (2), since $\theta \in\left[0, \frac{\pi}{2}\right]$, therefore
$$1 \leqslant \sin \theta+\cos \theta \leqslant \sqrt{2}$$

Also,
$$\frac{3+2 \sin \theta \cos \theta+\frac{1}{2}}{\sin \theta+\cos \theta}=(\sin \theta+\cos \theta)+\frac{5}{2} \cdot \frac{1}{\sin \theta+\cos \theta}$$

When $1 \leqslant x \leqslant \sqrt{2}$, $f(x)=x+\frac{5}{2} \cdot \frac{1}{x}$ is a decreasing function, so
$$\begin{aligned}
a & \geqslant \max _{\theta \in\left[0, \frac{\pi}{2}\right]} \frac{3+2 \sin \theta \cos \theta+\frac{1}{2}}{\sin \theta+\cos \theta} \\
& =1+\frac{5}{2}=\frac{7}{2}
\end{aligned}$$

Under condition (3), since
$$\begin{aligned}
\frac{3+2 \sin \theta \cos \theta-\frac{1}{2}}{\sin \theta+\cos \theta} & =(\sin \theta+\cos \theta)+\frac{3}{2} \cdot \frac{1}{\sin \theta+\cos \theta} \\
& \geqslant 2 \cdot \sqrt{\frac{3}{2}}=\sqrt{6}
\end{aligned}$$

and the equality holds when $\sin \theta+\cos \theta=\frac{\sqrt{6}}{2}$, thus
$$\min _{\theta \in\left[0, \frac{\pi}{2}\right]} \frac{3+2 \sin \theta \cos \theta-\frac{1}{2}}{\sin \theta+\cos \theta}=\sqrt{6}$$

Therefore, we have $a \leqslant \sqrt{6}$.
In summary, the range of $a$ is $a \geqslant \frac{7}{2}$ or $a \leqslant \sqrt{6}$.