Question

Can someone PLEASE help me with 4 and 5??? I’m legitimately struggling.

Answer 1

4) The key here is recognizing that the diagonals of a rectangle are congruent, and that they bisect each other. Therefore, AE = BE, and AE + BE = BD. Substituting, you get $(2x+3)+(12-x)=BD \\ x+15=BD$.
BD=x+7

5) The sides of a rectangle are parallel to each other. When you draw lines extending the sides, the implications of this become obvious. Suddenly, you have two parallel lines intersected by a transversal, and you probably know those angle relationships. BAC = ACD, so $3x+5=40-2x$. $x=7$.

To find AED (I believe there's an easier way, but I'm not sure), see that AED is the supplement of AEB, and AEB is 180° $- 2(BAC) = 180-6x-10 = 170-6(7)=128$. Therefore, AED is $180-128=52$. 
AED=52°

Answer 2

Problem 4.

In a rectangle, the diagonals are congruent and bisect each other.
You have AC = BD, and AE = CE = BE = DE.
Also each segment of a diagonal is half the length of the diagonal.

AE = BE

2x + 3 = 12 - x

3x = 9

x = 3

AE = 2x + 3 = 2(3) + 3 = 6 + 3 = 9

BD = 2(AE) = 2(9) = 18

Problem 5.

Sides AB and CD are parallel.
Angles BAC and ACD are alternate interior angles of parallel lines, so they are congruent.

m<BAC = m<ACD

3x + 5 = 40 - 2x

5x = 35

x = 7

m<ACD = 40 - 2x = 40 - 2(7) = 40 - 14 = 26

m<BDC = m<ACD = 26

<AED is a remote exterior angle to angles BDC and ACD.

m<AED = m<BCD + m<ACD = 26 + 26 = 52