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 following inequality always holds:
$$
(x+3+2 \sin \theta \cos \theta)^{2}+(x+a \sin \theta+a \cos \theta)^{2} \geqslant \frac{1}{8} .
$$
$\therefore$ The known inequality $\Leftrightarrow(3+2 \sin \theta \cos \theta-a \sin \theta$
$-a \cos \theta)^{2} \geqslant \frac{1}{4}$, 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 $0 \in\left[0, \frac{\pi}{2}\right]$;
or $a \leqslant \frac{3+2 \sin \theta \cos \theta-\frac{1}{2}}{\sin \theta+\cos \theta}$,
for any $0 \in\left[0, \frac{\pi}{2}\right]$. Under condition (2), since $0 \in\left[0, \frac{\pi}{2}\right]$,
$$
\begin{array}{l}
1 \leqslant \sin \theta+\cos \theta \leqslant \sqrt{2}, \\
\frac{3+2 \sin \theta \cos \theta+\frac{1}{2}}{\sin \theta+\cos \theta} \\
=(\sin \theta+\cos \theta)+\frac{5}{2} \cdot \frac{1}{\sin \theta+\cos \theta} .
\end{array}
$$

Clearly, when $1 \leqslant x \leqslant \sqrt{2}$, $f(x)=x+\frac{5}{2} \cdot \frac{1}{x}$ is a convex function. Therefore,
$$
\begin{array}{l}
\max _{\theta \in\left(0, \frac{\pi}{2}\right)} \frac{3+2 \sin \theta \cos \theta+\frac{1}{2}}{\sin \theta+\cos \theta} \\
=\max _{1<x<\sqrt{2}} f(x)=1+\frac{5}{2}=\frac{7}{2} .
\end{array}
$$

Thus, we get $a \geqslant \frac{7}{2}$.
Under condition (3), since
$$
\begin{array}{l}
\frac{3+2 \sin \theta \cos \theta-\frac{1}{2}}{\sin \theta+\cos \theta} \\
=(\sin \theta+\cos \theta)+\frac{3}{2} \cdot \frac{1}{\sin \theta+\cos \theta} \\
\geqslant 2 \sqrt{\frac{3}{2}}=\sqrt{6},
\end{array}
$$

and the equality holds when $\sin \theta+\cos \theta=\frac{\sqrt{6}}{2}$, thus, $\min _{0<\theta<\frac{\pi}{2}} \frac{3+2 \sin \theta \cos \theta-\frac{1}{2}}{\sin \theta+\cos \theta}=\sqrt{6}$.
Therefore, $a \leqslant \sqrt{6}$.
In summary, $a \geqslant \frac{7}{2}$ or $a \leqslant \sqrt{6}$ is the solution.