II. (This question is worth 25 points)
Find the range of real numbers $a$ such that for any real number $x$ and any $\theta \in\left[0, \frac{\pi}{2}\right]$, the inequality $(x+3+2 \sin \theta \cos \theta)^{2}+$ $(x+a \sin \theta+a \cos \theta)^{2} \geqslant \frac{1}{8}$ always holds.
$a \geqslant \frac{7}{2}$ or $a \leqslant \sqrt{6}$

II. 【Analysis and Solution】
For a given $\theta$, it is evident that when $x=\frac{3+2 \sin \theta \cos \theta+a \sin \theta+a \cos \theta}{2}$, the expression reaches its minimum value, which is $\left(\frac{3+2 \sin \theta \cos \theta+a \sin \theta+a \cos \theta}{2}\right)^{2} \times 2$. Therefore, the inequality is equivalent to
$(3+2 \sin \theta \cos \theta+a \sin \theta+a \cos \theta)^{2} \geqslant \frac{1}{4}$,
which expands to $3+2 \sin \theta \cos \theta+a \sin \theta+a \cos \theta \geqslant \frac{1}{2}$.
or $3+2 \sin \theta \cos \theta+a \sin \theta+a \cos \theta \leqslant-\frac{1}{2}$,
Under equation (1),
$a \geqslant \frac{3+2 \sin \theta \cos \theta+\frac{1}{2}}{\sin \theta+\cos \theta}, \theta \in\left[0, \frac{\pi}{2}\right]$.
And $\frac{3+2 \sin \theta \cos \theta+\frac{1}{2}}{\sin \theta+\cos \theta}=(\sin \theta+\cos \theta)+\frac{5}{2 \sin \theta+\cos \theta}$
When $\sin \theta+\cos \theta=1$, the above expression reaches its maximum value,
i.e., $\max \frac{3+2 \sin \theta \cos \theta+\frac{1}{2}}{\sin \theta+\cos \theta}=\frac{3+2+\frac{1}{2}}{1}=\frac{7}{2}$,
Thus, $a \geqslant \frac{7}{2}$.
Under equation (2),
$\frac{3+2 \sin \theta \cos \theta-\frac{1}{2}}{\sin \theta+\cos \theta}=(\sin \theta+\cos \theta)+\frac{3}{2 \sin \theta+\cos \theta}$,
its maximum value is reached when $\sin \theta+\cos \theta=\frac{\sqrt{3}}{2}$.
Thus, $a \leqslant \sqrt{6}$.
In summary, $a \geqslant \frac{7}{2}$ or $a \leqslant \sqrt{6}$.