13. Given the function
$$
f(x)=2 \sin ^{2}\left(\frac{\pi}{4}+x\right)-\sqrt{3} \cos 2 x-1,
$$

where, $x \in\left[\frac{\pi}{4}, \frac{\pi}{2}\right]$.
(1) Find the interval where $f(x)$ is monotonically increasing;
(2) If the inequality $|f(x)-m|<2$ holds for all $x \in$ $\left[\frac{\pi}{4}, \frac{\pi}{2}\right]$, find the range of the real number $m$.
Three, 13. (1) From the given,
$$
\begin{array}{l}
f(x)=2 \sin ^{2}\left(\frac{\pi}{4}+x\right)-\sqrt{3} \cos 2 x-1 \\
\Rightarrow f(x)=2 \sin \left(2 x-\frac{\pi}{3}\right)
\end{array}
$$

Then $-\frac{\pi}{2}+2 k \pi \leqslant 2 x-\frac{\pi}{3}$
$$
\leqslant \frac{\pi}{2}+2 k \pi\left(k \in \mathbf{Z} \text { and } x \in\left[\frac{\pi}{4}, \frac{\pi}{2}\right]\right) \text {. }
$$

Therefore, $x \in\left[\frac{\pi}{4}, \frac{5 \pi}{12}\right]$.
(2) From $|f(x)-m|<2$ being always true for $x \in\left[\frac{\pi}{4}, \frac{\pi}{2}\right]$, we have
$$
-2+m<f(x)<2+m \text {. }
$$

Also, $f(x)=2 \sin \left(2 x-\frac{\pi}{3}\right)\left(x \in\left[\frac{\pi}{4}, \frac{\pi}{2}\right]\right)$, so
$$
1 \leqslant f(x) \leqslant 2 \text {. }
$$

Therefore, $0<m<3$.