15. Let $y=f(x)$ be an odd function on $\mathbf{R}$, $f(x+2)=-f(x)$, and when $-1 \leqslant x \leqslant 1$, $f(x)=x^{3}$.
(1) Prove that the line $x=1$ is an axis of symmetry for the graph of the function $y=f(x)$.
(2) Find the analytical expression of the function $y=f(x)$ when $x \in[1,5]$.
(3) If $A=\{x|| f(x) |>a, x \in \mathbf{R}\}$, and $A \neq \varnothing$, find the range of the real number $a$.
15. (1) Since $f(1+x)=f[2+(x-1)]=-f(x-1)=f(1-x)$, i.e., $f(1+x)=f(1-x)$. Therefore, the graph of $y=f(x)$ is symmetric about the line $x=1$.
(2) Since when $-1 \leqslant x \leqslant 1$, $f(x)=x^{3}$.
When $x \in[1,3]$, $2-x \in[-1,1]$. Then $f(x)=f[1+(x-1)]=f[1-(x-1)]=f(2-x)=$ $-(x-2)^{3}$.
When $x \in[3,5]$, $x-2 \in[1,3]$, then $f(x)=-f(x-2)=(x-4)^{3}$.
Therefore, $f(x)=\left\{\begin{array}{ll}-(x-2)^{3} & x \in[1,3] \\ (x-4)^{3} & x \in[3,5] .\end{array}\right.$
(3) Since $f(x+4)=-f(x+2)=f(x)$. Hence $y=f(x)$ is a periodic function with a period of 4.
Therefore, $\{y \mid y=f(x), x \in \mathbf{R}\}=\{y \mid y=f(x), 1 \leqslant x \leqslant 5\}$.
Since $f(x)$ is a decreasing function on $[1,3]$. Therefore, on $[1,3]$, $-1 \leqslant f(x) \leqslant 1$. Also, $f(x)$ is an increasing function on $[3,5]$. Therefore, on $[3,5]$, $-1 \leqslant f(x) \leqslant 1$. Hence, the range of $f(x)$ is $[-1,1]$, i.e., $|f(x)| \leqslant 1$. To make $A \neq \varnothing$, we need $a<1$.