5. Let $f(x)$ be a function defined on $\mathbf{R}$ with a period of 2, which is even, strictly decreasing on the interval $[0,1]$, and satisfies $f(\pi)=1, f(2 \pi)=2$. Then the solution set of the inequality system $\left\{\begin{array}{l}1 \leq x \leq 2, \\ 1 \leq f(x) \leq 2\end{array}\right.$ is $\qquad$.
Solution: Given that $f(x)$ is an even function and strictly decreasing on $[0,1]$, we know that $f(x)$ is strictly increasing on $[-1,0]$. Combining this with the fact that $f(x)$ has a period of 2, we can conclude that $[1,2]$ is a strictly increasing interval for $f(x)$.
Notice that
$$
f(\pi-2)=f(\pi)=1, f(8-2 \pi)=f(-2 \pi)=f(2 \pi)=2,
$$

Therefore,
$$
1 \leq f(x) \leq 2 \Leftrightarrow f(\pi-2) \leq f(x) \leq f(8-2 \pi),
$$

Since $1<\pi-2<8-2 \pi<2$, the original inequality holds if and only if $x \in[\pi-2,8-2 \pi]$.