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
$$
\left\{\begin{array}{l}
1 \leqslant x \leqslant 2, \\
1 \leqslant f(x) \leqslant 2
\end{array}\right.
$$

is $\qquad$
5. $[\pi-2,8-2 \pi]$.

Given that $f(x)$ is an even function and strictly decreasing on the interval $[0,1]$, we know that $f(x)$ is strictly increasing on the interval $[-1,0]$. Combining this with the fact that $f(x)$ has a period of 2, we know that $[1,2]$ is a strictly increasing interval for $f(x)$.
Notice that,
$$
\begin{array}{l}
f(\pi-2)=f(\pi)=1, \\
f(8-2 \pi)=f(-2 \pi)=f(2 \pi)=2 .
\end{array}
$$

Then $1 \leqslant f(x) \leqslant 2$
$$
\Leftrightarrow f(\pi-2) \leqslant f(x) \leqslant f(8-2 \pi) \text {. }
$$

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