1. Given that the function $f(x)$ is a decreasing function on $\mathbf{R}$ and is an odd function. If $m, n$ satisfy
$$
\left\{\begin{array}{l}
f(m)+f(n-2) \leqslant 0, \\
f(m-n-1) \leqslant 0,
\end{array}\right.
$$

then the range of $5 m-n$ is
$-1 .[7,+\infty)$
Since $f(x)$ is an odd function and is defined at $x=0$, we have $f(0)=0$.
According to the problem, we have
$$
\left\{\begin{array}{l}
f(m) \leqslant -f(n-2)=f(2-n), \\
f(m-n-1) \leqslant 0=f(0) .
\end{array}\right.
$$

Since $f(x)$ is a decreasing function, it follows that
$$
\left\{\begin{array}{l}
m \geqslant 2 - n, \\
m - n - 1 \geqslant 0
\end{array} \Rightarrow \left\{\begin{array}{l}
m+n \geqslant 2, \\
m-n \geqslant 1 .
\end{array}\right.\right.
$$

Then $5 m-n=2(m+n)+3(m-n)$
$$
\geqslant 2 \times 2+3 \times 1=7 \text {, }
$$

Thus, the range of $5 m-n$ is $[7,+\infty)$.