Answer:
$$
\frac{21 \sqrt{3}}{8}
$$

Let segments $A I$ and $E F$ meet at $K$. Extending $A K$ to meet the circumcircle again at $Y$, we see that $X$ and $Y$ are diametrically opposite, and it follows that $A X$ and $E F$ are parallel. Therefore the height from $X$ to $\overline{U V}$ is merely $A K$. Observe that $A E=A F$, so $\triangle A E F$ is equilateral; since $M N, M P$ are parallel to $A F, A E$ respectively, it follows that $\triangle M V U, \triangle U E N, \triangle F P V$ are equilateral as well. Then $M V=M P-P V=\frac{1}{2} A C-F P=\frac{1}{2} A C-A F+A P=\frac{1}{2} A C-A F+\frac{1}{2} A B=\frac{1}{2} B C$, since $E, F$ are the tangency points of the incircle. Since $\triangle M V U$ is equilateral, we have $U V=M U=M V=\frac{1}{2} B C$.
Now we can compute $B C=7$, whence $U V=\frac{7}{2}$ and
$$
A K=\frac{A B+A C-B C}{2} \cdot \cos 30^{\circ}=\frac{3 \sqrt{3}}{2} .
$$

Hence, the answer is $\frac{21 \sqrt{3}}{8}$.
6. Let $A B C$ be a triangle with incenter $I$, incircle $\gamma$ and circumcircle $\Gamma$. Let $M, N, P$ be the midpoints of sides $\overline{B C}, \overline{C A}, \overline{A B}$ and let $E, F$ be the tangency points of $\gamma$ with $\overline{C A}$ and $\overline{A B}$, respectively. Let $U$, $V$ be the intersections of line $E F$ with line $M N$ and line $M P$, respectively, and let $X$ be the midpoint of arc $\widehat{B A C}$ of $\Gamma$. Given that $A B=5, A C=8$, and $\angle A=60^{\circ}$, compute the area of triangle $X U V$.