Example 5 (1) Find the smallest positive period of the function $f(x)=\sqrt[3]{\sin \frac{3}{2} x}-\sqrt[5]{\cos \frac{2}{3} x}$.
(2) Find the smallest positive period of the function $f(x)=2 \sin 3 x+3|\sin 4 x|$.
(1) For any $T(\neq 0), \sqrt[5]{\sin \frac{3}{2}(x+T)}=\sqrt[5]{\sin \frac{3}{2} x} \Leftrightarrow \sin \frac{3}{2}(x+T)=\sin \frac{3}{2} x$,

Therefore, the smallest positive period of the odd function $f_{1}(x)=\sqrt[3]{\sin \frac{3}{2} x}$ is $T_{1}=\frac{2 \pi}{3 / 2}=\frac{4}{3} \pi$, and the smallest positive period of the even function $f_{2}(x)=$ $-\sqrt[5]{\cos \frac{2}{3} x}$ is $T_{2}=\frac{2 \pi}{2 / 3}=3 \pi$, hence the smallest positive period of $f(x)$ is the least common multiple of $T_{1}$ and $T_{2}$, which is $12 \pi$.
(2) The smallest positive period of the odd function $f_{1}(x)=2 \sin 3 x$ is $T_{1}=\frac{2}{3} \pi$, and the smallest positive period of the even function $f_{2}(x)=3|\sin 4 x|$ is $T_{2}=\frac{\pi}{8}$, hence the smallest positive period of $f(x)$ is the least common multiple of $T_{1}$ and $T_{2}$, which is $2 \pi$.

Explanation: The above conclusion can not only be used to find the smallest positive period of the sum of odd and even functions, but also, through translation transformations, to find the smallest positive period of periodic functions that can be transformed into the sum of odd and even functions. For example:
Find the smallest positive period of $f(x)=(\sin x)^{2008}+(\cos x)^{20107}$.
Solution: $f\left(\frac{\pi}{2}-x\right)=(\cos x)^{2000}+(\sin x)^{2007}$, it is easy to see that $f_{1}(x)=(\sin x)^{2007}$ is an odd function, and $f_{1}(x+T)=f_{1}(x) \Leftrightarrow \sin (x+T)=\sin x$. Therefore, the smallest positive period of $f_{1}(x)$ is the same as that of $\sin x$, both being $T_{1}=$ $2 \pi$; $f_{2}(x)=(\cos x)^{20018}$ is an even function, and $f_{2}(x+T)=f_{2}(x) \Leftrightarrow|\cos (x+T)|=|\cos x|$, so the smallest positive period of $f_{2}(x)$ is the same as that of $|\cos x|$, both being $T_{2}=\pi$. Hence, the smallest positive period of $f\left(\frac{\pi}{2}-x\right)$ is the least common multiple of $T_{1}$ and $T_{2}$, which is $2 \pi$. Since the smallest positive period of $f(x)$ is the same as that of $f\left(\frac{\pi}{2}-x\right)$, the smallest positive period of $f(x)$ is $2 \pi$.

In general, if $f_{1}(x)$ represents one of the six basic trigonometric functions $(\sin x, \cos x, \tan x, \cot x, \sec x, \csc x)$, and $f(x)=a f_{1}^{2 n}(x)+b \cos ^{2 m-1} x\left(n, m \in \mathbf{N}^{*}, a b \neq 0\right)$, then the smallest positive period of $f(x)$ is $2 \pi$.

Note: When finding $T_{1}$ and $T_{2}$, it is essential to consider that the domain of $f_{1}(x)$ and $f_{2}(x)$ is the common domain $D$, otherwise the conclusion may not hold.

Counterexample: Let $n \in \mathbf{Z}$, the odd function $f_{1}(x)=\sin \pi x, x \neq 2 n$, and the even function $f_{2}(x)=\left\{\begin{array}{l}0, x \neq 4 n \\ 1, x=4 n\end{array}\right.$, then $T_{1}=2, T_{2}=4$, and the least common multiple of $T_{1}$ and $T_{2}$ is 4, but the smallest positive period of $f(x)=f_{1}(x)+f_{2}(x)=\sin \pi x, x \neq 2 n$ is $T=2$.

Why is this so? The reason lies in the fact that when finding $T_{1}$, the domain of $f_{1}(x)$ is $\mathbf{R}$, not the common domain $D=\{x \mid x \neq 2 n, n \in \mathbf{Z}, x \in \mathbf{R}\}$ of $f_{1}(x)$ and $f_{2}(x)$. In fact, the smallest positive period of $f_{1}(x)(x \in D)$ is $T_{1}=2$.