7. Let $f(x)$ be a function defined on $\mathbf{R}$ with a period of 2, which is even, and strictly decreasing on the interval $[1,2]$, and satisfies $f(\pi)=1, f(2 \pi)=0$. Then the solution set of the inequality system $\left\{\begin{array}{l}0 \leq x \leq 1, \\ 0 \leq f(x) \leq 1\end{array}\right.$ is
Answer: $[2 \pi-6,4-\pi]$.
Solution: Given that $f(x)$ is an even function and strictly decreasing on $[1,2]$, we know that $f(x)$ is strictly increasing on $[-2,-1]$. Combining this with the fact that $f(x)$ has a period of 2, we can conclude that $[0,1]$ is a strictly increasing interval for $f(x)$.
Notice that
$$
f(4-\pi)=f(\pi-4)=f(\pi)=1, f(2 \pi-6)=f(2 \pi)=0,
$$

Therefore,
$$
0 \leq f(x) \leq 1 \Leftrightarrow f(2 \pi-6) \leq f(x) \leq f(4-\pi),
$$

Since $0<2 \pi-6<4-\pi<1$, the original inequality holds if and only if $x \in[2 \pi-6,4-\pi]$.