14. Given $f(x)=\mathrm{e}^{x}-m x$.
(1) If $x>0$, the inequality
$$
(x-2) f(x)+m x^{2}+2>0
$$

always holds, find the range of the real number $m$;
(2) If $x_{1} 、 x_{2}$ are two zeros of the function $f(x)$, prove: $x_{1}+x_{2}>2$.
14. (1) Let $g(x)=(x-2) f(x)+m x^{2}+2$.

Then $g(x)=(x-2) \mathrm{e}^{x}+2 m x+2$.
By the problem, we know that when $x>0$, $g(x)>0$ always holds.
Notice,
$g^{\prime}(x)=(x-1) \mathrm{e}^{x}+2 m, g^{\prime \prime}(x)=x \mathrm{e}^{x}$.
Thus, when $x>0$, $g^{\prime \prime}(x)>0$, and $g^{\prime}(x)$ is an increasing function.
Also, from $g(2)=4 m+2>0$, we know $m>-\frac{1}{2}$.
(i) If $-\frac{1}{2}<m<\frac{1}{2}$, then $g^{\prime}(0)=-1+2 m<0, g^{\prime}(2)=\mathrm{e}^{2}+2 m>0$.
At this time, $g^{\prime}(x)$ has a unique zero point in the interval $(0,2)$, denoted as $x_{0}$. Then $0<x_{0}<2$.
Thus, when $0<x<x_{0}$, $g^{\prime}(x)<0$, and when $x>x_{0}$, $g^{\prime}(x)>0$.
Therefore, $g(x)$ is decreasing in the interval $(0, x_{0})$ and increasing in the interval $(x_{0}, +\infty)$.
Since $g(0)=0$, there exists $x \in (0, x_{0})$ such that $g(x)<0$, which contradicts the condition that $g(x)>0$ when $x>0$.
Thus, $-\frac{1}{2}<m<\frac{1}{2}$ does not meet the requirements.
(ii) If $m \geq \frac{1}{2}$, then $g^{\prime}(0)=-1+2 m \geq 0$.
At this time, $g(x)$ is an increasing function in the interval $[0,+\infty)$. Hence, when $x>0$, $g(x)>g(0)=0$.
Therefore, $m \geq \frac{1}{2}$ meets the requirements.
In summary, the range of $m$ is $\left[\frac{1}{2},+\infty\right)$.
(2) By the problem, we know $\mathrm{e}^{x_{1}}=m x_{1}, \mathrm{e}^{x_{2}}=m x_{2}$.
Then $m\left(x_{1}+x_{2}\right)=\mathrm{e}^{x_{1}}+\mathrm{e}^{x_{2}}$,
$m\left(x_{1}-x_{2}\right)=\mathrm{e}^{x_{1}}-\mathrm{e}^{x_{2}}$.
Assume without loss of generality that $x_{1}>x_{2}$.
It is easy to see that $m \neq 0$.
From equations (1) and (2), we get
$$
\begin{array}{l}
x_{1}+x_{2}=\frac{\left(x_{1}-x_{2}\right)\left(\mathrm{e}^{x_{1}}+\mathrm{e}^{x_{2}}\right)}{\mathrm{e}^{x_{1}}-\mathrm{e}^{x_{2}}} \\
=\frac{\left(x_{1}-x_{2}\right)\left(\mathrm{e}^{x_{1}-x_{2}}+1\right)}{\mathrm{e}^{x_{1}-x_{2}}-1} .
\end{array}
$$

For $m=\frac{1}{2}$, from (1) we know that when $x>0$,
$$
(x-2) \mathrm{e}^{x}+x+2>0
$$

always holds. At this time,
$$
\begin{array}{l}
\frac{x\left(\mathrm{e}^{x}+1\right)}{\mathrm{e}^{x}-1}-2=\frac{(x-2) \mathrm{e}^{x}+x+2}{\mathrm{e}^{x}-1}>0 . \\
\text { Hence } x_{1}+x_{2}=\frac{\left(x_{1}-x_{2}\right)\left(\mathrm{e}_{x_{1}-x_{2}}^{x_{1}}+1\right)}{\mathrm{e}^{x_{1}-x_{2}}-1}>2 .
\end{array}
$$