Solution:

Let $u = \sin^2 x$ and $v = \cos^2 x,$ so $u + v = 1,$ and set $p = uv.$ Expanding $(u+v)^5$ gives $u^5 + v^5 = 1 - 5p(u+v)^3 + 5p^2(u+v) = 1 - 5p + 5p^2,$ so the hypothesis reads $$1 - 5p + 5p^2 = \frac{11}{36} \quad\Longrightarrow\quad 36p^2 - 36p + 5 = 0,$$ with roots $p = \frac{1}{6}$ and $p = \frac{5}{6}.$ Since $p = \sin^2 x \cos^2 x = \frac{1}{4}\sin^2 2x \le \frac{1}{4},$ we must have $p = \frac{1}{6}.$

Similarly $u^6 + v^6 = (u^2 + v^2)^3 - 3p^2(u^2+v^2) = (1 - 2p)^3 - 3p^2(1 - 2p) = 1 - 6p + 9p^2 - 2p^3.$ Substituting $p = \frac{1}{6},$ $$u^6 + v^6 = 1 - 1 + \frac{1}{4} - \frac{1}{108} = \frac{26}{108} = \frac{13}{54},$$ so $m + n = 13 + 54 = 67.$