13. 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 $a$.
Three, 13. 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 $t \in[-1,1]$ and ensuring it is positive.

Since the function $f(t)=(t-a)^{2}+2 a-3$, we only need to discuss the three 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,
$$
\left\{\begin{array}{l}
a \leqslant-1, \\
f(-1)=a^{2}+4 a-2>0 .
\end{array}\right.
$$

Solving this, we get $a < -2 - \sqrt{6}$ or $a > -2 + \sqrt{6}$. Since $a \leqslant -1$, we have $a < -2 - \sqrt{6}$.
(2) When $-1 < a < 1$, the function $f(t)$ reaches its minimum value at $t=a$. Therefore,
$$
\left\{\begin{array}{l}
-1 < a < 1, \\
f(a)=2 a-3>0 .
\end{array}\right.
$$

Solving this, we get $a > \frac{3}{2}$. Since $-1 < a < 1$, there is no solution for $a$.
(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 this, we get $a > \sqrt{2}$.
Therefore, the range of $a$ that satisfies the condition is $a > \sqrt{2}$.