8. For the even function $y=f(x)$ with domain $R$, its graph is symmetric about the line $x=2$. It is known that when $x \in(-2,2)$, $f(x)=-x^{2}+1$. Then, on the interval $(4 k-2,4 k+2)(k \in \mathbf{Z})$, the analytical expression of $f(x)$ is $\qquad$.
8. $f(x)$ is a periodic function with a period of 4. When $x \in (4k-2, 4k+2)$, from $4k-2 < x < 4k+2$, we get $-2 < x-4k < 2$. When $x \in (-2, 2)$, $f(x) = -x^2 + 1$. Therefore, $f(x) = f(x-4k) = -(x-4k)^2 + 1 (k \in \mathbf{Z})$. This is the desired result.