Find the ratio of the area inside the square but outside the circle to the area of the square in the figure. All I'm worried about it is (a); for now. Okay, so let's start off like this: I know what the question is asking, sort of. I know it wants the ratio of the inside corner pieces of the square to the square itself. But what??? I obviously know that $A_{circle}=\pi r^2$ and $A_{square}=wh$ and also that the piece they are looking for is $A_{circle}-A_{square}$ but the ratio? I don't understand that. Can someone explain to me what they mean by that and stay posted for I may have questions about (b)-(e). Thanks.
If $A_0$ is the total area of the four corner pieces, and $A_1$ is the area of the square, they want $\frac{A_0}{A_1}$: that fraction is by definition the ratio of $A_0$ to $A_1$. Note also that for the square you have $w=h=2r$.
For Q a) 

The answer is $(4-\pi):4$ (or) $(4-\pi)r^2:4r^2$ 

First we have to find out the area of corner pieces. That we will get by subtracting the area of circle from the area of square. 
I.e.,area of square of side $2r$ is $4r^2$ ( bz. They gave half of the side is $r$)
And area of circle with radius $r$ is  $\pi r^2$
So area of corner pieces is $4r^2-\pi r^2 = (4-\pi) r^2$
Finally the ratio of area of corner pieces to square is

$$(4-\pi)r^2:4r^2$$  (or)
                                                     $$(4-pi):4$$