13. The function $f(x)$ defined on $\mathbf{R}$ satisfies: $f(x+2)=2-f(x), f(x+3) \geqslant f(x)$, try to find $f(x)$.
13. Since $f(x+3)=f[(x+1)+2]=2-f(x+1)$

Therefore, $2-f(x+1) \geqslant f(x)$, which means
$$
f(x+1)+f(x) \leqslant 2 \text {, }
$$

Also, $f(x)=2-f(x+2)$. Substituting into the right side of $f(x+3) \geqslant f(x)$, we get
$$
f(x+3) \geqslant 2-f(x+2),
$$

which implies $f(x+3)+f(x+2) \geqslant 2$.
Replacing $x$ with $x-2$, we get
$$
f(x+1)+f(x) \geqslant 2,
$$

From equations (1) and (2), we have $f(x+1)+f(x)=2$. Combining this with $f(x+2)=2-f(x)$, we get $f(x+2)=f(x+1)$. Replacing $x$ with $x-1$, we get $f(x+1)=f(x)$, so $f(x+2)=f(x)=2-f(x)$, which means $f(x)=1$.