Problem 813

Let $t=\cos x$, then the function $f(t)=t^{2}-2 a t+a^{2}+2 a-3$ has a positive minimum value on $t \in[-1,1]$.
(1) When $a \leqslant-1$, the function $f(t)$ is increasing on $t \in[-1,1]$, the minimum value is $f(-1)=$ $a^{2}+4 a-2>0$, solving this yields $a<0$, which is always true for $a \leqslant-1$.
(2) When $-1<a<1$, the function $f(t)$ reaches its minimum value at $t=a$, so $f(a)=a^{2}-2 a^{2}+a^{2}+2 a-3=2 a-3>0$, solving this yields $a>\frac{3}{2}$, which contradicts $-1<a<1$.
(3) When $a \geqslant 1$, the function $f(t)$ is decreasing on $t \in[-1,1]$, the minimum value is $f(1)=a^{2}-2 a+2 a-3+1=a^{2}-2>0$, solving this yields $a>\sqrt{2}$.
Therefore, the range of $a$ that satisfies the condition is $a\leqslant-1$ or $a>\sqrt{2}$.