2. 2. $25 \underset{\star}{\star \star}$ 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:
$$
(x+3+2 \sin \theta \cos \theta)^{2}+(x+a \sin \theta+a \cos \theta)^{2} \geqslant \frac{1}{8} .
$$
Obviously, the original problem is about the quadratic inequality in $x$: 
$$x^{2}+(3+2 \sin \theta \cos \theta+a \sin \theta+a \cos \theta) x+\frac{1}{2}(3+2 \sin \theta \cos \theta)^{2}+\frac{1}{2}(a \sin \theta+a \cos \theta)^{2}-\frac{1}{16} \geqslant 0$$ 
which always holds. Therefore, for $\theta \in \left[0, \frac{\pi}{2}\right]$, the discriminant $\Delta \leqslant 0$ always holds. That is, 
$$(3+2 \sin \theta \cos \theta-a \sin \theta-a \cos \theta)^{2} \geqslant \frac{1}{4}$$ 
for $\theta \in \left[0, \frac{\pi}{2}\right]$. From this, we get that for all $\theta \in \left[0, \frac{\pi}{2}\right]$,
$$
a \geqslant \frac{3+2 \sin \theta \cos \theta+\frac{1}{2}}{\sin \theta+\cos \theta} \cdots \text { (1), or } a \leqslant \frac{3+2 \sin \theta \cos \theta-\frac{1}{2}}{\sin \theta+\cos \theta} \cdots \text { (2). }
$$

Since $\theta \in \left[0, \frac{\pi}{2}\right]$, we have $1 \leqslant \sin \theta+\cos \theta=\sqrt{2} \sin \left(\theta+\frac{\pi}{4}\right) \leqslant \sqrt{2}$.
From (1), we have $a \geqslant \sin \theta+\cos \theta+\frac{5}{2} \cdot \frac{1}{\sin \theta+\cos \theta}$. It is easy to see that when $1 \leqslant x \leqslant \sqrt{2}$, $f(x) = x+\frac{5}{2} \cdot \frac{1}{x}$ is a decreasing function. Therefore, 
$$\max _{\left(\theta \in \left[0, \frac{\pi}{2}\right]\right)}\left(\sin \theta+\cos \theta+\frac{5}{2} \cdot \frac{1}{\sin \theta+\cos \theta}\right) = \max _{1 \leqslant x \leqslant \sqrt{2}} f(x) = 1+\frac{5}{2} = \frac{7}{2}.$$ 
Thus, we get $a \geqslant \frac{7}{2}$.

From (2), we have $a \leqslant \sin \theta+\cos \theta+\frac{3}{2} \cdot \frac{1}{\sin \theta+\cos \theta}$. And 
$$\sin \theta+\cos \theta+\frac{3}{2} \cdot \frac{1}{\sin \theta+\cos \theta} \geqslant 2 \sqrt{\frac{3}{2}} = \sqrt{6},$$ 
and the equality holds when $\sin \theta+\cos \theta=\frac{\sqrt{6}}{2}$. Therefore, we get $a \leqslant \sqrt{6}$.
In summary, $a \geqslant \frac{7}{2}$ or $a \leqslant \sqrt{6}$ is the solution.