13. Let the function be
$$
f(x)=4 \sin x \cdot \sin ^{2}\left(\frac{\pi}{4}+\frac{x}{2}\right)+\cos 2 x \text {. }
$$

If $|f(x)-m|<2$ holds for $\frac{\pi}{6} \leqslant x \leqslant \frac{2 \pi}{3}$, then the range of the real number $m$ is
13. $(1,4)$.

Notice
$$
\begin{array}{l}
f(x)=2 \sin x\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^{2}+\cos 2 x \\
=2 \sin x(1+\sin x)+1-2 \sin ^{2} x \\
=1+2 \sin x . \\
\text { When } \frac{\pi}{6} \leqslant x \leqslant \frac{2 \pi}{3} \text {, }|f(x)-m|<2 \text { always holds }
\end{array}
$$

When $\frac{\pi}{6} \leqslant x \leqslant \frac{2 \pi}{3}$, $|f(x)-m|<2$ always holds, i.e., $f(x)-2<m<f(x)+2$ always holds.
Then $(f(x)-2)_{\text {max }}<m<(f(x)+2)_{\text {min }}$.
It is easy to find that $(f(x))_{\max }=3,(f(x))_{\min }=2$.
Therefore, $1<m<4$.