8. Given an even function $y=f(x)$ defined on $\mathbf{R}$ that satisfies $f(x)=f(2-x)$, when $x \in[0,1]$, $f(x)=x^{2}$, then for $2 k-1 \leqslant x \leqslant 2 k+1(k \in \mathbf{Z})$, the expression for $f(x)$ is $\qquad$.
8. This problem must deduce a period of the function using the given conditions to optimize the solution.

For any $x \in \mathbf{R}$, we have $f(x+2)=f[2-(-x)]=f(-x)=f(x)$, so $f(x)$ is a periodic function with a period of 2. Since $f(x)$ is an even function, when $x \in [0,1]$, $f(x)=x^{2}$. Therefore, when $x \in [-1,1]$, $f(x)=x^{2}$. When $2k-1 \leqslant x \leqslant 2k+1$ $(k \in \mathbf{Z})$, we have $-1 \leqslant x-2k \leqslant 1$. Since $f(x)$ is a periodic function with a period of 2, we have $f(x)=f(x-2k)=(x-2k)^{2}$. Therefore, when $2k-1 \leqslant x \leqslant 2k+1$ $(k \in \mathbf{Z})$, $f(x)=(x-2k)^{2}$.