14. As shown in the figure, $F_{1}, F_{2}$ are the left and right foci of the hyperbola $C: \frac{x^{2}}{4}-y^{2}=1$. A moving point $P\left(x_{0}, y_{0}\right)\left(y_{0} \geqslant 1\right)$ is on the right branch of the hyperbola $C$. Let the angle bisector of $\angle F_{1} P F_{2}$ intersect the $x$-axis at point $M(m, 0)$ and the $y$-axis at point $N$.
（1）Find the range of $m$;
(2) Let the line $l$ passing through $F_{1}, N$ intersect the hyperbola $C$ at points $D, E$. Find the maximum value of the area of $\triangle F_{2} D E$.
(1) $\frac{\left|F_{1} M\right|}{\left|M F_{2}\right|}=\frac{\left|P F_{1}\right|}{\left|P F_{2}\right|}=\frac{e x_{0}+a}{e x_{0}-a}$
$\Rightarrow \frac{m+\sqrt{5}}{\sqrt{5}-m}=\frac{\frac{\sqrt{5}}{2} x_{0}+2}{\frac{\sqrt{5}}{2} x_{0}-2}$
$$
=\frac{\sqrt{5}+\frac{4}{x_{0}}}{\sqrt{5}-\frac{4}{x_{0}}} \Rightarrow m=\frac{4}{x_{0}} \text {. }
$$

Thus, $k_{P M}=\frac{y_{0}}{x_{0}-\frac{4}{x_{0}}}=\frac{x_{0} y_{0}}{x_{0}^{2}-4}=\frac{x_{0}}{4 y_{0}} \Rightarrow P M: y=\frac{x_{0}}{4 y_{0}}\left(x-\frac{4}{x_{0}}\right)=\frac{x_{0} x}{4 y_{0}}-\frac{1}{y_{0}}$ $\Rightarrow \frac{x_{0} x}{4}-y_{0} y=1$. From $y_{0}^{2}=\frac{x_{0}^{2}}{4}-1 \geqslant 1 \Rightarrow x_{0} \geqslant 2 \sqrt{2} \Rightarrow m=\frac{4}{x_{0}} \in(0, \sqrt{2}]$.
(2) In the line $P M$, let $x=0 \Rightarrow y_{N}=-\frac{1}{y_{0}} \Rightarrow F_{1} N: x=-\sqrt{5} y_{0} y-\sqrt{5}$.

Solving $\left\{\begin{array}{l}x=-\sqrt{5} y_{0} y-\sqrt{5}, \\ x^{2}-4 y^{2}=4\end{array} \Rightarrow\left(5 y_{0}^{2}-4\right) y^{2}+10 y_{0} y+1=0\right.$.
Let $D\left(x_{1}, y_{1}\right), E\left(x_{2}, y_{2}\right)$, then $\left|y_{E}\right|-\left|y_{D}\right|=\left|y_{1}-y_{2}\right|=\frac{\sqrt{100 y_{0}^{2}-4\left(5 y_{0}^{2}-4\right)}}{5 y_{0}^{2}-4}=\frac{4 \sqrt{5 y_{0}^{2}+1}}{5 y_{0}^{2}-4}$ $\Rightarrow S_{\triangle F_{2} D E}=\frac{1}{2} \cdot 2 \sqrt{5} \cdot \frac{4 \sqrt{5 y_{0}^{2}+1}}{5 y_{0}^{2}-4} \xlongequal{t=\sqrt{5 y_{0}^{2}+1} \geqslant \sqrt{6}} \frac{4 \sqrt{5} t}{t^{2}-5}=\frac{4 \sqrt{5}}{t-\frac{5}{t}} \leqslant \frac{4 \sqrt{5}}{\sqrt{6}-\frac{5}{\sqrt{6}}}=4 \sqrt{30}$. Therefore, the maximum area of $\triangle F_{2} D E$ is $4 \sqrt{30}$.