21. Let the function $y=f(x)$ be defined on $\mathbf{R}$, for any real numbers $m, n$, it always holds that $f(m+n)=f(m) \cdot f(n)$, and when $x>0$, $0<f(x)<1$;
(2) Find the range of $x$ that satisfies $f\left(2 x^{2}-4 x-1\right), f(x-1)<1$.
21. (1) Take $m>0$, then $00$, we have $01$.
(2) Take any $x_{1}, x_{2}$ in $\mathbf{R}$, and $x_{1}1$, so $f\left(x_{1}-x_{2}\right)-1>0$, and $f\left(x_{2}\right)>0$, thus $f\left(x_{1}\right)-f\left(x_{2}\right)>0$. That is, $f\left(x_{1}\right)>f\left(x_{2}\right)$, hence $f(x)$ is a decreasing function on $\mathbf{R}$.
Since $f(0)=1$. Therefore, $f\left(2 x^{2}-4 x-1\right) \cdot f(x-1)0$, solving this gives $x2$.