14. (25 points) Given the functions
$$
f(x)=x-\ln x-2, \quad g(x)=x \ln x+x \text {. }
$$
(1) Prove that $f(x)$ has a zero in the interval $(3,4)$;
(2) If $k \in \mathbf{Z}$, and $g(x)>k(x-1)$ for all $x>1$, find the maximum value of $k$.
14. (1) Let $f(x)=x-\ln x-2$. Then
$$
f^{\prime}(x)=1-\frac{1}{x} \text {. }
$$

Therefore, $f(x)$ is monotonically increasing on the interval $(1,+\infty)$.
$$
\text { Also, } f(3)=1-\ln 30 \text {, }
$$

Thus, $f(x)$ has a zero point $x_{0} \in(3,4)$.
(2) The unique zero point $x_{0}$ of $f(x)$ clearly satisfies
$$
x_{0}-\ln x_{0}-2=0 \text {, }
$$

and when $x \in\left(1, x_{0}\right)$, $f(x)f\left(x_{0}\right)=0$;
when $x>1$,
$$
g(x)>k(x-1) \Leftrightarrow \frac{x \ln x+x}{x-1}>k .
$$

Let $h(x)=\frac{x \ln x+x}{x-1}$. Then
$$
h^{\prime}(x)=\frac{x-\ln x-2}{(x-1)^{2}}=\frac{f(x)}{(x-1)^{2}} \text {. }
$$

Thus, $h^{\prime}(x)$ has the same sign as $f(x)$.
Therefore, when $x \in\left(1, x_{0}\right)$, $h^{\prime}(x)0$.
Hence, $h(x)$ is monotonically decreasing on the interval $\left(1, x_{0}\right)$ and monotonically increasing on the interval $\left(x_{0},+\infty\right)$.
Therefore, $h(x)_{\min }=h\left(x_{0}\right)=\frac{x_{0}\left(\ln x_{0}+1\right)}{x_{0}-1}$
$$
=\frac{x_{0}\left(x_{0}-1\right)}{x_{0}-1}=x_{0} \text {. }
$$

By the problem, we know $k<h(x)_{\min }=x_{0}(k \in \mathbf{Z})$, and $x_{0} \in(3,4)$.
Therefore, the maximum value of $k$ is 3.