2. Let $t=\cos x$, then the original inequality becomes
$$
t^{2}-2 a t+a^{2}+2 a-3>0, t \in[-1,1] \text {. }
$$

Thus, the problem is transformed into finding the minimum value of the function $f(t)=t^{2}-2 a t+a^{2}+2 a-3$ on $t \in[-1,1]$ to be positive.
Since the function $f(t)=(t-a)^{2}+2 a-3$, we only need to discuss the 3 possible positions of the axis of symmetry $t=a$ relative to the interval $[-1,1]$.
(1) When $a \leqslant-1$, the function $f(t)$ is increasing on $t \in[-1,1]$, so the minimum value is $f(-1)$. Therefore,
$$
\left\{\begin{array}{l}
a \leqslant-1, \\
f(-1)=a^{2}+4 a-2>0 .
\end{array} \text { Solving gives } a0 .
\end{array}\right.
$$

In this case, there is no value of $a$.
(3) When $a \geqslant 1$, the function $f(t)$ is decreasing on $t \in[-1,1]$, so the minimum value is $f(1)$. Therefore, $\left\{\begin{array}{l}a \geqslant 1, \\ f(1)=a^{2}-2>0 .\end{array}\right.$ Solving gives $a>\sqrt{2}$.
Thus, the range of $a$ that satisfies the condition is $a>\sqrt{2}$.
2. (2003 Tianjin Competition Question) The solution set of the inequality $a^{2}+2 a-\sin ^{2} x-2 a \cos x>2$ about $x$ is all real numbers. Find the range of real numbers for $a$.

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