Answer: 222

Small Hint:

Rewrite the first equation as $\sin^2 A + \sin^2 B$ $- 2\sin A \sin B \cos C = \frac{1}{8};$ the laws of sines and cosines turn the left side into $\sin^2 C$

Big Hint:

So $\sin^2 C = \frac{1}{8}$ and $\sin^2 A = \frac{4}{9},$ and the requested quantity is $2 - \sin^2 B$ with $\sin B = \sin(A + C)$

Solution:

Replacing each $\cos^2$ by $1 - \sin^2,$ the first equation becomes $\sin^2 A + \sin^2 B$ $- 2 \sin A \sin B \cos C = \frac{1}{8}.$ By the law of sines, $\sin A = \frac{a}{2R}$ and so on, so the left side equals $$\begin{aligned} \frac{a^2 + b^2 - 2ab\cos C}{4R^2} &= \frac{c^2}{4R^2} \\ &= \sin^2 C \end{aligned}$$ by the law of cosines. Hence $\sin^2 C = 2 - \frac{15}{8} = \frac{1}{8}.$ The same argument turns the second equation into $\sin^2 A = 2 - \frac{14}{9} = \frac{4}{9},$ and shows the requested expression equals $2 - \sin^2 B.$

Since $A$ and $C$ are acute, $\cos A = \frac{\sqrt{5}}{3}$ and $\cos C = \frac{\sqrt{14}}{4},$ with $\sin A = \frac{2}{3}$ and $\sin C = \frac{\sqrt{2}}{4}.$ Then $$\begin{aligned} \sin B &= \sin(A + C) \\ &= \frac{2}{3} \cdot \frac{\sqrt{14}}{4} + \frac{\sqrt{5}}{3} \cdot \frac{\sqrt{2}}{4} \\ &= \frac{2\sqrt{14} + \sqrt{10}}{12}, \end{aligned}$$ so $\sin^2 B = \frac{66 + 8\sqrt{35}}{144} = \frac{33 + 4\sqrt{35}}{72}.$

Therefore $2 - \sin^2 B = \frac{111 - 4\sqrt{35}}{72},$ and $p + q + r + s$ $= 111 + 4 + 35 + 72$ $= 222.$