22. (12 points) Let the function $f(x)=\frac{x}{\ln x}-a x$.
(1) If the function $f(x)$ is monotonically decreasing on the interval $(1,+\infty)$, find the minimum value of the real number $a$;
(2) If there exist $x_{1}, x_{2} \in\left[\mathrm{e}, \mathrm{e}^{2}\right]$, satisfying
$$
f\left(x_{1}\right) \leqslant f^{\prime}\left(x_{2}\right)+a,
$$

find the range of the real number $a$.
22. (1) From the given, we have $x>0, x \neq 1$.

Since $f(x)$ is a decreasing function on the interval $(1,+\infty)$, it follows that, on the interval $(1,+\infty)$, we always have
$$
\begin{array}{l} 
f^{\prime}(x)=\frac{\ln x-1}{\ln ^{2} x}-a \leqslant 0 . \\
\text { Also, } f^{\prime}(x)=\frac{\ln x-1}{\ln ^{2} x}-a \\
=-\left(\frac{1}{\ln x}-\frac{1}{2}\right)^{2}+\frac{1}{4}-a,
\end{array}
$$

Thus, when $\frac{1}{\ln x}=\frac{1}{2}$, i.e., $x=\mathrm{e}^{2}$,
$$
f^{\prime}(x)_{\max }=\frac{1}{4}-a \leqslant 0 .
$$

Therefore, $a \geqslant \frac{1}{4}$, which means the minimum value of $a$ is $\frac{1}{4}$.
(2) The statement “there exist $x_{1}, x_{2} \in\left[\mathrm{e}, \mathrm{e}^{2}\right]$ such that
$$
f\left(x_{1}\right) \leqslant f^{\prime}\left(x_{2}\right)+a "
$$

is equivalent to “when $x \in\left[\mathrm{e}, \mathrm{e}^{2}\right]$, we have
$$
f(x)_{\text {min }} \leqslant f^{\prime}(x)_{\text {max }}+a ” .
$$

From (1), when $x \in\left[\mathrm{e}, \mathrm{e}^{2}\right]$,
$$
f^{\prime}(x)_{\max }=\frac{1}{4}-a \Rightarrow f^{\prime}(x)_{\max }+a=\frac{1}{4} \text {. }
$$

The problem is equivalent to “when $x \in\left[\mathrm{e}, \mathrm{e}^{2}\right]$, we have $f(x)_{\min } \leqslant \frac{1}{4}$.
(i) When $a \geqslant \frac{1}{4}$, from (1), $f(x)$ is a decreasing function on the interval $\left[e, e^{2}\right]$, then
$$
\begin{array}{l}
f(x)_{\text {min }}=f\left(\mathrm{e}^{2}\right)=\frac{\mathrm{e}^{2}}{2}-a \mathrm{e}^{2} \leqslant \frac{1}{4} \\
\Rightarrow a \geqslant \frac{1}{2}-\frac{1}{4 \mathrm{e}^{2}} .
\end{array}
$$
(ii) When $a<\frac{1}{4}$, if $-a<0$, then $f^{\prime}(x)<0$, and $f(x)$ is a decreasing function.
$$
\begin{array}{l}
\text { Then } f(x)_{\text {min }}=f\left(\mathrm{e}^{2}\right)=\frac{\mathrm{e}^{2}}{2}-a \mathrm{e}^{2} \leqslant \frac{1}{4} \\
\Rightarrow a \geqslant \frac{1}{2}-\frac{1}{4 \mathrm{e}^{2}}>\frac{1}{4}, \text { a contradiction. }
\end{array}
$$
If $-a>0$, then $f^{\prime}(x)>0$, and $f(x)$ is an increasing function.
$$
\begin{array}{l}
\text { Then } f(x)_{\text {min }}=f\left(x_{0}\right)=\frac{x_{0}}{\ln x_{0}}-a x_{0} \\
\leqslant \frac{1}{4}\left(x_{0} \in\left(\mathrm{e}, \mathrm{e}^{2}\right)\right) \\
\Rightarrow a \geqslant \frac{1}{\ln x_{0}}-\frac{1}{4 x_{0}}>\frac{1}{\ln \mathrm{e}^{2}}-\frac{1}{4 \mathrm{e}} \\
>\frac{1}{2}-\frac{1}{4}=\frac{1}{4},
\end{array}
$$

which contradicts $0<a<\frac{1}{4}$.
In summary, $a \geqslant \frac{1}{2}-\frac{1}{4 \mathrm{e}^{2}}$.