5. Given that the point $(a, b)$ moves on the curve $\arcsin x=\arccos y$, and the ellipse $a x^{2}+b y^{2}=1$ is outside the circle $x^{2}+y^{2}=$ $\frac{2}{\sqrt{3}}$ (including the case where they are tangent). Then, the range of $\arcsin b$ is $\qquad$ .
5. $\left[\frac{\pi}{6}, \frac{\pi}{4}\right) \cup\left(\frac{\pi}{4}, \frac{\pi}{3}\right]$.

From $\arcsin x=\arccos y$, we get
$$
y=\sqrt{1-x^{2}} \quad(0 \leqslant x \leqslant 1) .
$$

The semi-major and semi-minor axes of the ellipse $\frac{x^{2}}{\frac{1}{a}}+\frac{y^{2}}{\frac{1}{b}}=1$ are $\frac{1}{\sqrt{a}}$ and $\frac{1}{\sqrt{b}}$ respectively. According to the problem, we should have
$$
\frac{1}{\sqrt{a}} \geqslant \sqrt{\frac{2}{\sqrt{3}}} \text { and } \frac{1}{\sqrt{b}} \geqslant \sqrt{\frac{2}{\sqrt{3}}}
$$

and $b=\sqrt{1-a^{2}} \quad(0<a<1)$.
Solving this, we get $\frac{1}{2} \leqslant a \leqslant \frac{\sqrt{3}}{2}, \frac{1}{2} \leqslant b \leqslant \frac{\sqrt{3}}{2}$.

Furthermore, since $a \neq b$, we have $b \neq \frac{\sqrt{2}}{2}$.
Thus, $\frac{\pi}{6} \leqslant \arcsin b<\frac{\pi}{4}$ or $\frac{\pi}{4}<\arcsin b \leqslant \frac{\pi}{3}$.