22. Given the function $f(x)=a x-1-\ln x(a \in \mathbf{R})$.
(1) Discuss the number of extremum points of $f(x)$ within its domain;
(2) If the function $f(x)$ has an extremum at $x=1$, and for any $x \in(0,+\infty), f(x) \geqslant b x-2$ always holds, find the range of the real number $b$;
(3) When $x>y>\mathrm{e}-1$, prove: $\mathrm{e}^{x-y}>\frac{\ln (x+1)}{\ln (y+1)}$.
(1) $f^{\prime}(x)=a-\frac{1}{x}(x>0), a \leqslant 0$ when, $f(x)$ is monotonically decreasing on $(0,+\infty)$, with no extremum points; when $a>0$, $f(x)$ is monotonically decreasing on $\left(0, \frac{1}{a}\right)$ and monotonically increasing on $\left(\frac{1}{a},+\infty\right)$, so $x=\frac{1}{a}$ is the minimum point of $f(x)$, and $f(x)$ has no maximum point.
(2) From (1), we know $f(x)=x-1-\ln x$, then $f(x) \geqslant b x-2 \Rightarrow b \leqslant \frac{x+1-\ln x}{x}$,

Let $g(x)=\frac{x+1-\ln x}{x}, g^{\prime}(x)=\frac{-2+\ln x}{x^{2}}$, thus $g(x)$ is monotonically decreasing on $\left(0, \mathrm{e}^{2}\right)$ and monotonically increasing on $\left(\mathrm{e}^{2},+\infty\right)$, hence $g(x)_{\min }=g\left(\mathrm{e}^{2}\right)=\frac{\mathrm{e}^{2}-1}{\mathrm{e}^{2}}$.

Therefore, the range of real number $b$ is $\left(-\infty, \frac{\mathrm{e}^{2}-1}{\mathrm{e}^{2}}\right]$.
(3) The original inequality is equivalent to $\frac{\mathrm{e}^{x}}{\ln (x+1)}>\frac{\mathrm{e}^{y}}{\ln (y+1)} \Leftrightarrow \frac{\ln (x+1)}{\mathrm{e}^{x}}y>\mathrm{e}-1 \Rightarrow h(x)<h(y)$, so the original inequality holds.