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
$$
a \geqslant \frac{3+2 \sin \theta \cos \theta+\frac{1}{2}}{\sin \theta+\cos \theta} \text {, for any } \theta \in\left[0, \frac{\pi}{2}\right]
$$

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

and $\frac{3+2 \sin \theta \cos \theta+\frac{1}{2}}{\sin \theta+\cos \theta}=(\sin \theta+\cos \theta)+\frac{5}{2} \cdot \frac{1}{\sin \theta+\cos \theta}$,
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 $\sin \theta+\cos \theta=\frac{\sqrt{6}}{2}$, therefore
$$
\min _{\theta \in\left[0 . \frac{\pi}{2}\right]} \frac{3+2 \sin \theta \cos \theta-\frac{1}{2}}{\sin \theta+\cos \theta}=\sqrt{6},
$$

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