Blue area

From the diagram on the left, you can see that the blue area is the area of the square minus 4 of the shaded region on the left. Therefore, we want to find the area of the shaded region.

From the diagram on the right, we know that:

Shaded area = Area of Sector AED - Area of Segment AE

Furthermore, we know that:

Area of Segment AE = Area of Sector BAE (with center B) - Area of Triangle ABE

Now, let the sides of the square (and the radius of the circles) be $a$

We know that AE is the radius of Sector ABED, therefore AE $=a$

We know that BE is the radius of Sector BAEC, therefore BE $=a$

Therefore, triangle ABE is an equilateral triangle with sides $a$

We can then calculate everything out now:

Area of Segment AE

= Area of Sector BAE (with center B) - Area of Triangle ABE

$= \dfrac{60}{360}\pi a^2 - \dfrac{1}{2}a^2 \sin 60^{\circ}\\ =\dfrac{\pi}{6}a^2 - \dfrac{\sqrt{3}}{4}a^2\\ =\dfrac{2\pi - 3\sqrt{3}}{12}a^2$

Area of shaded region

= Area of Sector AED - Area of Segment AE

$=\dfrac{30}{360}\pi a^2 - \dfrac{2\pi - 3\sqrt{3}}{12}a^2 \\ =\dfrac{\pi}{12}a^2 - \dfrac{2\pi - 3\sqrt{3}}{12}a^2\\ =\dfrac{3\sqrt{3} - \pi}{12}a^2$

Blue area (shared area)

= Area of square - 4 $\times$ Area of shaded region

$=a^2 - 4 \times \dfrac{3\sqrt{3} - \pi}{12}a^2\\ =\dfrac{3 + \pi -3\sqrt{3}}{3}a^2$

The ratio of shared area to area of square

$=\dfrac{\left(\dfrac{3 + \pi -3\sqrt{3}}{3}a^2\right)}{a^2}\\ =\boxed{\dfrac{\pi +3 -3\sqrt{3}}{3}}$

The sketch explains how to find the ratio for one fourth. For the hole too it will have the same ratio.

There is a similar problem here !