Let $ABC$ be a triangle with $AB = 5$, $BC = 7$, and $CA = 8$, and let $D$ be a point on arc $\widehat{BC}$ of its circumcircle $\Omega$. Suppose that the angle bisectors of $\angle ADB$ and $\angle ADC$ meet $AB$ and $AC$ at $E$ and $F$, respectively, and that $EF$ and $BC$ meet at $G$. Line $GD$ meets $\Omega$ at $T$. If the maximum possible value of $AT^2$ can be expressed as $\frac{a}{b}$ for positive integers $a, b$ with $\gcd (a,b) = 1$, find $a + b$.

[i]Proposed by Andrew Wu[/i]
1. **Identify the key elements and setup the problem:**
   - Given triangle \(ABC\) with sides \(AB = 5\), \(BC = 7\), and \(CA = 8\).
   - Point \(D\) is on the arc \(\widehat{BC}\) of the circumcircle \(\Omega\).
   - Angle bisectors of \(\angle ADB\) and \(\angle ADC\) meet \(AB\) and \(AC\) at \(E\) and \(F\) respectively.
   - \(EF\) and \(BC\) intersect at \(G\).
   - Line \(GD\) meets \(\Omega\) at \(T\).
   - We need to find the maximum possible value of \(AT^2\).

2. **Prove that \(T\) is the midpoint of arc \(\widehat{BC}\):**
   - Let \(M\) and \(N\) be the midpoints of arcs \(\widehat{AB}\) and \(\widehat{AC}\) respectively.
   - Let \(I\) be the incenter of \(\triangle ABC\).
   - By Pascal's Theorem on hexagon \(ABNDMC\), points \(E\), \(I\), and \(F\) are collinear.
   - If \(TD\) and \(BC\) intersect at \(G\), then by Pascal's Theorem on hexagon \(TDMCBA\), points \(G\), \(E\), and \(I\) are collinear.
   - Since \(E\), \(I\), and \(F\) are collinear, \(G\) lies on line \(EF\), proving that \(T\) is the midpoint of arc \(\widehat{BC}\).

3. **Calculate \(AT\) using the properties of the circumcircle:**
   - Given \(\angle BAC = 60^\circ\), we have \(\angle BTC = 120^\circ\) because \(T\) is the midpoint of arc \(\widehat{BC}\).
   - Since \(BC = 7\) and \(BT = CT\), we use the property of the circumcircle to find \(BT\):
     \[
     BT = CT = \frac{BC}{2 \sin(60^\circ)} = \frac{7}{2 \cdot \frac{\sqrt{3}}{2}} = \frac{7}{\sqrt{3}}
     \]
   - By Ptolemy's Theorem for cyclic quadrilateral \(ABCT\):
     \[
     AB \cdot CT + AC \cdot BT = AT \cdot BC
     \]
     Substituting the known values:
     \[
     5 \cdot \frac{7}{\sqrt{3}} + 8 \cdot \frac{7}{\sqrt{3}} = AT \cdot 7
     \]
     Simplifying:
     \[
     \frac{35}{\sqrt{3}} + \frac{56}{\sqrt{3}} = 7 \cdot AT
     \]
     \[
     \frac{91}{\sqrt{3}} = 7 \cdot AT
     \]
     \[
     AT = \frac{91}{7\sqrt{3}} = \frac{13}{\sqrt{3}}
     \]
   - Therefore, \(AT^2\) is:
     \[
     AT^2 = \left(\frac{13}{\sqrt{3}}\right)^2 = \frac{169}{3}
     \]

4. **Conclusion:**
   - The maximum possible value of \(AT^2\) is \(\frac{169}{3}\).
   - Therefore, \(a = 169\) and \(b = 3\), and \(a + b = 172\).

The final answer is \(\boxed{172}\).