Solution:

By the angle bisector theorem, $AE = \frac{13}{27} \cdot 15$ and $CD = \frac{14}{29} \cdot 13,$ so $E$ lies on $\overline{AN}$ and $D$ lies on $\overline{MC},$ with $$NE = AN - AE = \frac{15}{2} - \frac{65}{9} = \frac{5}{18}, \qquad MD = CM - CD = \frac{13}{2} - \frac{182}{29} = \frac{13}{58}.$$

Since $AMPN$ is cyclic, $\angle ENP = \angle ANP = 180^\circ - \angle AMP = \angle DMP,$ and since $AEPD$ is cyclic, $\angle NEP = 180^\circ - \angle AEP = \angle ADP = \angle MDP.$ Hence triangles $ENP$ and $DMP$ are similar, so $\frac{NP}{MP} = \frac{NE}{MD}.$ By the law of sines in triangles $ANP$ and $AMP,$ whose angles $\angle ANP$ and $\angle AMP$ are supplementary, $$\frac{\sin\angle BAQ}{\sin\angle CAQ} = \frac{\sin\angle NAP}{\sin\angle MAP} = \frac{NP}{MP} = \frac{5/18}{13/58} = \frac{145}{117}.$$

Comparing the areas of triangles $ABQ$ and $ACQ,$ which share the cevian $\overline{AQ},$ $$\frac{BQ}{CQ} = \frac{[ABQ]}{[ACQ]} = \frac{AB \sin\angle BAQ}{AC \sin\angle CAQ} = \frac{15}{13} \cdot \frac{145}{117} = \frac{725}{507},$$ which is in lowest terms since $507 = 3 \cdot 13^2$ and $725 = 5^2 \cdot 29.$ Thus $m - n = 725 - 507 = 218.$