Let $\Gamma$ be the circumcircle of a triangle $ABC$ and let $E$ and $F$ be the intersections of the bisectors of $\angle ABC$ and $\angle ACB$ with $\Gamma$. If $EF$ is tangent to the incircle $\gamma$ of $\triangle ABC$, then find the value of $\angle BAC$.
1. **Identify the given elements and their properties:**
   - $\Gamma$ is the circumcircle of $\triangle ABC$.
   - $E$ and $F$ are the intersections of the angle bisectors of $\angle ABC$ and $\angle ACB$ with $\Gamma$.
   - $EF$ is tangent to the incircle $\gamma$ of $\triangle ABC$.
   - We need to find the value of $\angle BAC$.

2. **Use the properties of angle bisectors and the circumcircle:**
   - The angle bisectors of $\angle ABC$ and $\angle ACB$ intersect $\Gamma$ at points $E$ and $F$ respectively.
   - Since $E$ and $F$ lie on the circumcircle, $\angle BAE = \angle CAE$ and $\angle CAF = \angle BAF$.

3. **Consider the tangency condition:**
   - $EF$ is tangent to the incircle $\gamma$ of $\triangle ABC$.
   - The tangency condition implies that $EF$ is perpendicular to the line segment joining the incenter $I$ to the point of tangency.

4. **Use the fact that $EF$ bisects $AI$:**
   - Since $EF$ is tangent to the incircle and perpendicular to $AI$, it must bisect $AI$.
   - This implies that $AI$ is the diameter of the circle with $EF$ as the tangent.

5. **Relate the inradius $r$ to the angle $\angle BAC$:**
   - Let the inradius be $r$.
   - Since $AI$ is perpendicular to $EF$ and $EF$ bisects $AI$, we have $AI = 2r$.
   - The angle $\angle BAC$ can be expressed in terms of the angles at $E$ and $F$.

6. **Calculate $\angle BAC$:**
   - Since $AI$ is the diameter, $\angle BAI = \angle CAI = 30^\circ$.
   - Therefore, $\angle BAC = 2 \times \angle BAI = 2 \times 30^\circ = 60^\circ$.

\[
\boxed{60^\circ}
\]