12. (14 points) Given the function
$$
\begin{array}{l}
f(x)=\ln x-a x(a>0), \\
g(x)=f(x)+f^{\prime}(x) .
\end{array}
$$
(1) If the maximum value of the function $f(x)$ is -4 when $1 \leqslant x \leqslant \mathrm{e}$, find the expression for the function $f(x)$;
(2) Determine the range of values for $a$ such that the function $g(x)$ is monotonic on the interval $(0,+\infty)$.
12. (1) Notice that, $f^{\prime}(x)=\frac{1}{x}-a$.

Thus, $f(x)$ is monotonically increasing in $\left(0, \frac{1}{a}\right)$, and $f(x)$ is monotonically decreasing in $\left(\frac{1}{a},+\infty\right)$.
Therefore, when $x=\frac{1}{a}$, $f(x)$ reaches its maximum value.
(i) When $01$,
$f(x)_{\max }=f(1)=-4$.
Solving this gives $a=4$.
(ii) When $1 \leqslant \frac{1}{a} \leqslant \mathrm{e}$, i.e., $\frac{1}{\mathrm{e}} \leqslant a \leqslant 1$, $f(x)_{\max }=f\left(\frac{1}{a}\right)=-4$.
Solving this gives $a=\mathrm{e}^{3}>1$ (discard).
(iii) When $\frac{1}{a}>e$, i.e., $0\frac{1}{\mathrm{e}}$ (discard).
In summary, $f(x)=\ln x-4 x$.
(2) From the problem, we know
$g(x)=\ln x-a x+\frac{1}{x}-a$.
Then $g^{\prime}(x)=\frac{1}{x}-a-\frac{1}{x^{2}}$
$=-\left(\frac{1}{x}-\frac{1}{2}\right)^{2}+\frac{1}{4}-a$.
(i) $a \geqslant \frac{1}{4}$.
Then $g^{\prime}(x) \leqslant 0$, and only when $x=2, a=\frac{1}{4}$, $g^{\prime}(x)=0$. Therefore, $g(x)$ is monotonically decreasing in $(0,+\infty)$.
(ii) $00$;
If $\frac{1}{x} \in\left(\frac{1}{2}+\sqrt{\frac{1}{4}-a},+\infty\right)$, then $g^{\prime}(x)<0$. Therefore, $g(x)$ is not monotonically increasing in $(0,+\infty)$.
In summary, the range of $a$ is $\left[\frac{1}{4},+\infty\right)$.