Triangle $ABC$ is inscribed in circle $\omega$ with $AB=5$, $BC=7$, and $AC=3$. The bisector of angle $A$ meets side $\overline{BC}$ at $D$ and circle $\omega$ at a second point $E$. Let $\gamma$ be the circle with diameter $\overline{DE}$. Circles $\omega$ and $\gamma$ meet at $E$ and a second point $F$. Then $AF^2 = \frac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
Use the angle bisector theorem to find $CD=\tfrac{21}{8}$, $BD=\tfrac{35}{8}$, and use Stewart's Theorem to find $AD=\tfrac{15}{8}$. Use Power of Point $D$ to find $DE=\tfrac{49}{8}$, and so $AE=8$. Use law of cosines to find $\angle CAD = \tfrac{\pi} {3}$, hence $\angle BAD = \tfrac{\pi}{3}$ as well, and $\triangle BCE$ is equilateral, so $BC=CE=BE=7$.

In triangle $AEF$, let $X$ be the foot of the altitude from $A$; then $EF=EX+XF$, where we use signed lengths. Writing $EX=AE \cdot \cos \angle AEF$ and $XF=AF \cdot \cos \angle AFE$, we get
\begin{align}\tag{1}     EF = AE \cdot \cos \angle AEF + AF \cdot \cos \angle AFE. \end{align}
Note $\angle AFE = \angle ACE$, and the Law of Cosines in  $\triangle ACE$ gives  $\cos \angle ACE = -\tfrac 17$. 
Also, $\angle AEF = \angle DEF$, and $\angle DFE = \tfrac{\pi}{2}$ ($DE$ is a diameter), so $\cos \angle AEF = \tfrac{EF}{DE} =  \tfrac{8}{49}\cdot EF$.
Plugging in all our values into equation $(1)$, we get:
\[EF = \tfrac{64}{49} EF -\tfrac{1}{7} AF \quad \Longrightarrow \quad EF = \tfrac{7}{15} AF.\]
The Law of Cosines in  $\triangle AEF$, with $EF=\tfrac 7{15}AF$ and $\cos\angle AFE = -\tfrac 17$ gives
\[8^2 = AF^2 + \tfrac{49}{225}  AF^2 + \tfrac 2{15} AF^2 = \tfrac{225+49+30}{225}\cdot AF^2\]
Thus $AF^2 = \frac{900}{19}$. The answer is $\boxed{919}$.