"2.Where did they even get 2sa+sb=0 from?"
From the assumption that the length of the rope does not change.
"Why did they determine the change in distance this way? My first assumption was that if block B moved up 2ft, then block A should move down 2ft (the rope must "move" 2ft too right?)."
No, there is a difference between a movable pulley and a fixed pulley, so A moves down 1 ft.
"3.Where did 2vA=−vB come from?"
From the same assumption as above. However, the direction of vA is shown incorrectly (or, alternatively, the signs in the formula are wrong).
"4 The FBD for block A is confusing, why is the friction force FA in the direction of the ropes? I thought it was block B that is going down?"
It is writen in the statement of the problem that block B went up.
" Am I the only one who had trouble deducting that the pulley and block A are the same object?"
I don't quite understand what this phrase means exactly and how it is relevant.
"5.Look at the final answer, how could vb be negative? The problem says block B goes UP."
See the answer to your item 3.
Hint : Friction opposes tendency to move. Tension is produced if the string is stretched $very$ slightly. So, increase friction to maximum and then tension will act if necessary.
Ironically, you are thinking absolutely right. Give yourself a cookie.
From part $a$, we know that the blocks will be at rest at all angles below that.
You are also right as at very small angles there is no need of tension and we can ignore it to solve for, again, an angle condition. You have done excellent work. Congrats.
Now we come to the middle angles. Oh... they drive you insane, don't they?
Let's start. We can start our analysis from 2 blocks, 1 will give a contradiction and other will give a result, but I will start with the one giving contradiction. This will help you.
All angles are in degrees :
$\theta=35 $
Lets start by analysing Block A (No racism intended)
Gravity is trying to pull it down : $5*10*\sin(35)N=28.67N$
Friction comes to the rescue(up) : $50*\cos(35)N=16.38N$ // read my hint to know why friction is put max here
As it is at rest, $T=12.29N$
Now Block B is also at rest,
Weight = $114.71N$
max f= $81.92N$
$16.38+114.71=12.29+f$
$f=118.8N$
OOPS, it exceed max value. So, Lets start by analysing Block B. (I love alliteration)
Gravity trying : $114.71N$
Friction comes to the rescue(up) : $81.92N$
You can take from here I guess. calculate tension. Note that you have to revise your calculation for tension again as reaction friction force will be provided by A. Better assume it $f$ from starting FBD of B.
This will yield the correct answer. Friction will be less than max value for upper block. In most cases, You should start analysing with heavier block(my experience). Hope your doubts are cleared.
Best Answer
The first thing first -The concept of friction Friction come in play when there is any tendency of relative motion between 2 surfaces and it is in opposite direction of relative motion (Note- I used word relative motion not simply motion). In simple word friction try to reduce relative motion. in 2 block problem and this problem the difference is T (tension). In both problem direction of relative motion is similar, friction is trying to do same thing i.e. pulling the other block with the one on whom force is applied so that relative motion can be reduced. But in this problem Tension forces block to move in other direction then supplied by friction.And but obvious T>Friction. so here T is increasing relative motion and friction as always trying to make them move together.