Let $f(x)=y$, then
$$
\begin{aligned}
& \sin \left(g(x)-\frac{\pi}{3}\right)=\frac{f(x)}{2} \sqrt{4-f^2(x)}=\frac{y}{2} \sqrt{4-y^2}=t \text { (let) } \\
& \Rightarrow \quad y^2-\frac{y^4}{4}=t^2 \quad \Rightarrow \quad 4 y^2-y^4=4 t^2 \\
& \Rightarrow \quad\left(y^2-2^2=-4 t^2+4 \quad \Rightarrow \quad t^2=1-\frac{1}{4}\left(y^2-2^2\right)^2\right. \\
& \because \quad f(x)=y \in[-1,1] \quad \Rightarrow \quad y^2 \in[0,1] \\
& \therefore \quad t^2 \in\left[0, \frac{3}{4}\right] \quad \Rightarrow \quad t \in\left[-\frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}\right] \\
& \text { So, } \sin \left(g(x)-\frac{\pi}{3}\right) \in\left[-\frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}\right] \\
& \Rightarrow \quad g(x)-\frac{\pi}{3} \in\left[-\frac{\pi}{3}, \frac{\pi}{3}\right] \Rightarrow g(x) \in\left[0, \frac{2 \pi}{3}\right]
\end{aligned}
$$