What's The Angle?

Let $\overline{OP} = r \implies \overline{OQ} = r - 1$ and let $\overline{OM} = r, m\angle{PMQ} = \alpha$ and $m\angle{QMO} = \beta$ .

$\tan(\beta) = \dfrac{r - 1}{r}$ and $\alpha + \beta = 45^{\circ} \implies \alpha = 45^{\circ} - \beta$

$\implies \tan(\alpha) = \tan(45^{\circ} - \beta) = \dfrac{1 - \tan(\beta)}{1 + \tan(\beta)} = \dfrac{1}{2r - 1}$

and

$A_{R} = r^2 - \dfrac{\pi}{4}r^2 = \dfrac{4 - \pi}{4}r^2 = \dfrac{\pi^2}{8}$ $\implies r = \dfrac{\pi}{\sqrt{2(4 - \pi)}} \implies$

$\tan(\beta) = \dfrac{\pi - \sqrt{2(4 - \pi)}}{\pi} \implies \tan(\alpha) = \dfrac{\sqrt{2(4 - \pi)}}{2\pi - \sqrt{2(4 - \pi)}}$

$\implies \alpha = \arctan(\dfrac{\sqrt{2(4 - \pi)}}{2\pi - \sqrt{2(4 - \pi)}} ) \approx \boxed{14.760926556951^{\circ}}$ .

Let the radius of the circle be $r$ . Then the side length of the square is $2r$ and its area $A_\square = 4r^2$ and $A_R = \dfrac {A_\square - \pi r^2}4 = \dfrac {(4-\pi)r}4 = \dfrac {\pi^2}8 \implies r = \dfrac \pi{\sqrt{8-2\pi}} \approx 2.3976631135$ . Then

$\angle PMQ = \angle PMO - \angle QMO = 45^\circ - \tan^{-1} \frac {r-1}r \approx 45^\circ - 30.239073443^\circ \approx \boxed{14.8}^\circ$