14. Given $f(x)=e^{x}-m x$.
(1) If $x>0$, the inequality $(x-2) f(x)+m x^{2}+2>0$ always holds, find the range of real number $m$;
(2) If $x_{1}, x_{2}$ are two zeros of the function $f(x)$, prove that $x_{1}+x_{2}=2$.
(1) Let $g(x)=(x-2) f(x)+m x^{2}+2$, then $g(x)=(x-2) e^{x}+2 m x+2$. For $x>0$, the inequality $(x-2) f(x)+m x^{2}+2>0$ always holds $\Leftrightarrow$ for $x>0$, $g(x)>0$ always holds.

Since
$$
g^{\prime}(x)=e^{x}+(x-2) e^{x}+2 m=(x-1) e^{x}+2 m, g^{\prime \prime}(x)=e^{x}+(x-1) e^{x}=x e^{x}
$$

Therefore, for $x>0$, $g^{\prime \prime}(x)=x e^{x}>0$, and $g^{\prime}(x)=(x-1) e^{x}+2 m$ is an increasing function on the interval $[0,+\infty)$.
Also, from $g(2)=4 m+2$, we know $m>-\frac{1}{2}$.
(1) If $-\frac{1}{2}<m<\frac{1}{2}$, then
$$
g^{\prime}(0)=-1+2 m<0, \lim _{x \rightarrow+\infty} g^{\prime}(x)=+\infty
$$

At this time, $g^{\prime}(x)$ has a unique zero point in the interval $(0,2)$, denoted as $x_{0}$, then $0<x_{0}<2$, and $g^{\prime}(x)<0$ for $x \in(0, x_{0})$, $g^{\prime}(x)>0$ for $x \in\left(x_{0},+\infty\right)$, so $g(x)$ is decreasing on $(0, x_{0})$ and increasing on $\left(x_{0},+\infty\right)$, and $g(0)=0$, so there exists $x \in(0, x_{0})$ such that $g(x)<0$, which contradicts the condition that $g(x)>0$ for $x>0$.
(2) If $m \geq \frac{1}{2}$, then for $x>0$,
$$
g^{\prime}(x)>g^{\prime}(0)=-1+2 m \geq 0
$$

At this time, $g(x)$ is an increasing function on $[0,+\infty)$, so for $x>0$, $g(x)>g(0)=0$, therefore, $m \geq \frac{1}{2}$ meets the requirement.
From (1) and (2), the range of $m$ is $\left[\frac{1}{2},+\infty\right)$;
(2) Since $x_{1}, x_{2}$ are the two zeros of the function $f(x)=e^{x}-m x$, we have
$$
e^{x_{1}}=m x_{1}, e^{x_{2}}=m x_{2}, m\left(x_{1}+x_{2}\right)=e^{x_{1}}+e^{x_{2}}, m\left(x_{1}-x_{2}\right)=e^{x_{1}}-e^{x_{2}}
$$

Assume $x_{1}>x_{2}$, it is easy to know $m \neq 0$, combining the above two equations and eliminating $m$, we get
$$
x_{1}+x_{2}=\frac{\left(x_{1}-x_{2}\right)\left(e^{x_{1}}+e^{x_{2}}\right)}{e^{x_{1}}-e^{x_{2}}}=\frac{\left(x_{1}-x_{2}\right)\left(e^{x_{1}-x_{2}}+1\right)}{e^{x_{1}-x_{2}}-1}
$$

Also, from (1), we know that for $m=\frac{1}{2}$, when $x>0$, $g(x)=(x-2) e^{x}+2 m x+2>0$ always holds.
So when $x>0$, $(x-2) e^{x}+x+2>0$ always holds; so when $x>0$,
$$
\frac{x\left(e^{x}+1\right)}{e^{x}-1}-2=\frac{x e^{x}+x-2 e^{x}+2}{e^{x}-1}=\frac{(x-2) e^{x}+x+2}{e^{x}-1}>0
$$

So
$$
x_{1}+x_{2}-2=\frac{\left(x_{1}-x_{2}\right)\left(e^{x_{1}-x_{2}}+1\right)}{e^{x_{1}-x_{2}}-1}-2>0, x_{1}+x_{2}>2
$$

When $x_{1}>2$, so $x_{1}+x_{2}>2$.
(20 points)