In the first one, it seems to me that angle c is part of a linear pair with 80 degrees, so it must be supplementary to 80 degrees.
And angle a is an alternate interior angle to the 47 degree angle and we know that the alternate interior angles are congruent.
On the second parallelogram, remember that the diagonals bisect each other.
3y + 13 = 2y + 37
Solve the above one for y, and substitute it into this one.
3y = 2x - 4
In the kite, triangle FEH is isosceles and since the vertex angle FEH = 60, it is also equilateral. This means FH = EF = EH = 16.
Also, FL = HL = 8 because EG bisects FH.
Using the 30-60-90 rule, you can easily find EL = . Subtract that from 20 to get LG. Then use the pythagorean theorem to find FG.
I think I've answered your last question, but angle EFG does not equal 60 degrees. Angel EFH = 60 degrees. Using trig, it's somewhere around 97.5 degrees, but that's not important here.