$\begin{aligned} & \int(\sqrt{1-\sin x}+\sqrt{1+\sin x}) d x \\ & \because \frac{3 \pi}{2} < x < \frac{5 \pi}{2} \Rightarrow \frac{3 \pi}{4} < \frac{x}{2} < \frac{5 \pi}{4} \\ & \because \ln \left(\frac{3 \pi}{4}, \frac{5 \pi}{4}\right), \sin \frac{x}{2}>\cos \frac{x}{2} \\ = & \int\left(\sqrt{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)^2}+\sqrt{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)^2}\right) d x \\ = & \left.\int\left(\left|\cos \frac{x}{2}-\sin \frac{x}{2}\right|+\cos \frac{x}{2}+\sin \frac{x}{2}\right)\right) d x \\ = & \int\left(\sin \frac{x}{2}-\cos \frac{x}{2}+\sin \frac{x}{2}+\cos \frac{x}{2}\right) d x \\ = & 2 \int \sin \frac{x}{2} d x=2 \times 2\left[-\cos \frac{x}{2}\right]+C \\ = & -4 \cos \frac{x}{2}+C \Rightarrow f(x)=-4 \cos \frac{x}{2} \\ \therefore & f\left(\frac{\pi}{3}\right)-f(0)=-4 \cos \left(\frac{\pi}{6}\right)+4 \cos 0=4-\frac{4 \sqrt{3}}{2} .\end{aligned}$