What's the measure of the angle?

Let the radius of the circle be $r$ . Then the side length of the square is $2r$ . And the area of the blue regions is:

$(2r)^2 - \pi r^2 = \frac {\pi^2}2 \implies r^2 = \frac {\pi^2}{2(4-\pi)} \implies r = \frac \pi{\sqrt{2(4-\pi)}} \approx 2.3976631135$

Therefore

$\alpha = 45^\circ - \tan^{-1} \left(\frac {r-1}r \right) \approx 45^\circ - 30.239073443^\circ \approx \boxed{14.8}$

Let $A_{T}$ be the sum of the blue shaded regions

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

$r = \dfrac{\pi}{\sqrt{2(4 - \pi)}}$ and $\alpha + \beta = 45^{\circ} \implies \alpha = 45^{\circ} - \beta \implies \theta = 90 + \beta$ and

$m = \sqrt{2}r = \dfrac{\pi}{\sqrt{4 - \pi}}$

Using law of sines on $\triangle{ABC} \implies \dfrac{\sin(45^{\circ} - \beta)}{1} = \dfrac{\sin(90^{\circ} + \beta)}{m} \implies$

$\dfrac{\pi}{\sqrt{4 - \pi}}\sin(45^{\circ} - \beta) = \sin(90^{\circ} + \beta) \implies$

$\pi\cos(\beta) - \pi\sin(\beta) = \sqrt{2(4 - \pi)}\cos(\beta) \implies$

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

$\beta = \arctan(\dfrac{\pi - \sqrt{2(4 - \pi)}}{\pi}) \implies \alpha = 45^{\circ} - \arctan(\dfrac{\pi - \sqrt{2(4 - \pi)}}{\pi}) \approx \boxed{14.760926556951^{\circ}}$ .

The area of the blue region, ${{A}_{blue}}$ , can be expressed as
$\text{Area of square }-\text{ Area of circle}$ , thus

${{A}_{blue}}={{\left( 2R \right)}^{2}}-\pi {{R}^{2}}\Rightarrow 4{{R}^{2}}-\pi {{R}^{2}}=\dfrac{{{\pi }^{2}}}{2}\Rightarrow {{R}^{2}}=\dfrac{{{\pi }^{2}}}{8-2\pi }\Rightarrow R=\dfrac{\pi }{\sqrt{8-2\pi }}$ On the right $\triangle OAB$ ,

$\begin{aligned} \tan \left( \angle CAO \right)=\dfrac{CO}{AO} & \Rightarrow \tan \left( 45{}^\circ -a \right)=\dfrac{R-1}{R} \\ & \Rightarrow \dfrac{1-\tan a}{1+\tan a}=1-\dfrac{1}{R} \\ & \Rightarrow \tan a=\dfrac{1}{2R-1}=\dfrac{1}{2\dfrac{\pi }{\sqrt{8-2\pi }}-1}=\dfrac{\sqrt{8-2\pi }}{2\pi -\sqrt{8-2\pi }} \\ \end{aligned}$ Hence,

$a={{\tan }^{-1}}\left( \frac{\sqrt{8-2\pi }}{2\pi -\sqrt{8-2\pi }} \right)\approx \boxed{14.76{}^\circ}$