16. (14 points) Given the function $f(x)=\frac{\ln x}{x}$.
(1) Suppose a real number $k$ makes $f(x)<k x$ always true, find the range of $k$.
(2) Let $g(x)=f(x)-k x(k \in \mathbf{R})$. If the function $g(x)$ has two zeros in the interval $\left[\frac{1}{\mathrm{e}}, \mathrm{e}^{2}\right]$, find the range of $k$.
16. (1) Let $h(x)=\frac{f(x)}{x}=\frac{\ln x}{x^{2}}(x>0)$.

Then $h^{\prime}(x)=\frac{1-2 \ln x}{x^{3}}(x>0)$.
Let $h^{\prime}(x)=\frac{1-2 \ln x}{x^{3}}=0 \Rightarrow x=\sqrt{\mathrm{e}}$.
When $x$ varies in the interval $(0,+\infty)$, the changes in $h^{\prime}(x)$ and $h(x)$ are as shown in Table 2.
Table 2
\begin{tabular}{|c|c|c|c|}
\hline$x$ & $(0, \sqrt{\mathrm{e}})$ & $\sqrt{\mathrm{e}}$ & $(\sqrt{\mathrm{e}},+\infty)$ \\
\hline$h^{\prime}(x)$ & + & 0 & - \\
\hline$h(x)$ & Increase & $\frac{1}{2 \mathrm{e}}$ & Decrease \\
\hline
\end{tabular}

From Table 2, we know that when $x=\sqrt{\mathrm{e}}$, $h(x)$ reaches its maximum value $\frac{1}{2 \mathrm{e}}$.
Given that for any $x>0$, it is always true that $k>\frac{f(x)}{x}=h(x)$.
Therefore, the range of $k$ is $\left(\frac{1}{2 \mathrm{e}},+\infty\right)$.
(2) Let $g(x)=0$ to get $k=\frac{f(x)}{x}=\frac{\ln x}{x^{2}}$.
From Table 2, we know that $h(x)=\frac{\ln x}{x^{2}}$ is an increasing function in the interval $\left[\frac{1}{\mathrm{e}}, \sqrt{\mathrm{e}}\right]$ and a decreasing function in the interval $\left[\sqrt{\mathrm{e}}, \mathrm{e}^{2}\right]$.
Note that,
$$
h\left(\frac{1}{\mathrm{e}}\right)=-\mathrm{e}^{2}, h(\sqrt{\mathrm{e}})=\frac{1}{2 \mathrm{e}}, h\left(\mathrm{e}^{2}\right)=\frac{2}{\mathrm{e}^{4}} .
$$

When $\frac{2}{\mathrm{e}^{4}} \leqslant k<\frac{1}{2 \mathrm{e}}$, the function $g(x)$ has two zeros in the interval $\left[\frac{1}{\mathrm{e}}, \mathrm{e}^{2}\right]$.
Thus, the range of $k$ is $\frac{2}{\mathrm{e}^{4}} \leqslant k<\frac{1}{2 \mathrm{e}}$.