A couple of questions, individual answers are also greatly appreciated!

Mmmmm....let's see..

1) Since BAC  = 23, then angle DAQ  will be complementary to this  = 90 - 23  = 67

And QDA  =  90 - DAQ  =  23

And since DP = PC, then in triangle DPC.....angle  DCP = angle PDC

But......since tranversal AC  cuts parallels DC and AB....then angle BAC  = angle DCP  = 23

So....PDC  also = 23

So....

QDA  + PDC   =  23 + 23   =  46°

3) 

Because transversal EF cuts parallels AB, CD.....then angle  BEF  = angle GFE

And since  EF = FG.....then angles FEG  and EGF are equal

So.....angle FEG  =  [  180  - (100 +x) ] / 2   =  (80 - x)/2

And angles  BEF +  FEG + AEG  = 180

Sustituting....we have   that

(100 + x)  +  (80 - x) / 2  +  x   =   180      multiply through by 2

2(100 + x)  + 80 - x  + 2x  =  360     simplify

200 +2x  + 80 - x + 2x  =  360

280  + 3x  =  360     subtract 280  from both sides

3x  =  80       divide both sides by 3

x  = 80/3 °   ≈  26.66°

4)  This one isn't too difficult

The total area of the rectangle  is  7 x 5   =  35 units^2    (1) 

The white triangle is obtuse.....

The base of this triangle......sitting on the "bottom" of the rectangle  = 2 units

The altitude occurs outside the triangle   and is  =  5 units

So....the area of this triangle is   (1/2) base * altitude  =  (1/2) (2) (5)  =  5 units^2    (2)

So.....the red  area is just the difference of (1)  and (2) =

35  -  5    =     30 units^2

P.S.   -  hope you're feeling better  !!!

2) Note that the side of the equilateral triangle = Side of the square

So.....draw EC

And in triangle ECB,  EB  = CB    and angle CBE  =  30°

But  ....since  EB  = CB, the angles opposite those sides, ECB  and CEB will be equal

And either will be  =     (180  - 30)  /  2   =   150 / 2   =  75°

So....angle ECB  =  75°

So.....angle  ECD  will be complementary to this  =   15°