Example 5 Given $f(x)=a \tan ^{3} x+b \sin x+1$, and $f(4)=5$, find the value of $f(2 \pi-4)$.
Analysis This problem is not a simple evaluation problem; it also requires finding a period of the function and determining its parity.
Solution Let $g(x)=a \tan ^{3} x+b \sin x$,
so $g(2 \pi+x)=a \tan ^{3}(2 \pi+x)+b \sin (2 \pi+x)=g(x)$,
thus, a period of $g(x)$ is $T=2 \pi$.
Also, since $g(-x)=a \tan ^{3}(-x)+b \sin (-x)=-g(x)$,
$g(x)$ is an odd function, i.e., $g(2 \pi-4)=g(-4)=-g(4)$,
so $f(x)=g(x)+1$, hence $5=g(4)+1$, i.e., $g(4)=4$, thus $g(2 \pi-4)=-4$,
therefore $f(2 \pi-4)=g(2 \pi-4)+1=-4+1=-3$.
!Comment The geometric characteristic of a periodic function is that the function graph in each period is the same, so first draw the graph within one period, then translate the graph by integer multiples of the period, to obtain the graph in the required range, and finally use the graph to solve the problem with the help of the idea of combining numbers and shapes, which is both intuitive and simple.

Alternative Solution Since $f(x)$ is a function with a period of 2. So when $k \in \mathbf{Z}$, $2 k$ is a period of $f(x)$, and when $x \in I_{k}$, $x-2 k \in I_{1}$. Therefore, $f(x)=(x-2 k)^{2}$, i.e., for $k \in \mathbf{Z}, x \in I_{k}$, $f(x)=(x-2 k)^{2}$.
(2) To find $M_{k}=\left\{a \mid\right.$ such that the equation $f(x)=a x$ has two distinct real roots in $I_{k}$, we can first concretize the equation $f(x)=a x$ to $(x-2 k)^{2}=a x$. Thus, the problem is transformed into finding the range of values for $a$ such that the equation has two distinct real roots in $(2 k-1,2 k+1]$. To determine the range of the slope $a$ of the linear function $y=a x$, we need to find the conditions under which the graphs of the two functions $y=a x$ and $y=(x-2 k)^{2}$ intersect at two different points in the interval $(2 k-1,2 k+1]$.