Working on parts 2 & 3, but I figured I would post part 1. In your drawing, LK and BC are parallel, making ALK = ABC and AKL = ACB. Therefore, ALK is similar to ABC. Since these triangles are similar:

(1)

If the side length of the inscribed square above side a is s, then the equation above says:

(2)

Now we draw the height of the triangle down from vertex A to side a, and call it h, meeting at the point H on a. Let the intersection of AH and LK be D. Line LK splits AH into two segments, the upper (AD) which has length h-s and the lower (DH) which has length s. Then, we now know that ADK is similar to AHC. Therefore,

(3)

And,

(4)

Thus, by equations (2), (3), and (4),

(5)

And then, solving for s, gives:

(6)

Could you explain exactly what you mean in part 3. of your question a little more? Thanks.