How complicated do you want this answer to be?
Answered simply:
1) Yes the length of the tower is a factor. More accurately, the way weight is distributed along the length is a factor.
If I understand you problem there is a tower (let's think telephone pole) laying flat on the ground with on end connected to a hinge (free to pivot but connected to the ground.) We're going to lift from the other end, and the question really is, "How much force does it take to lift the end of the pole?"
The solution lies in equating the moments (torques) created by the weight of the pole and the lift of the helicopter. If the tower (pole) is uniform in weight along its' length (so that it would balance if picked up at mid-length) then its' center of gravity is at the midpoint of its length, and that is the simple way to determine where the weight should be applied for this condition. (If the tower is heavier in one section than another then the center of gravity needs to be found).
The moment due to weight is the weight (times gravity to get force) times the distance from the center of gravity to the hinge. (Let's assume length/2.)
The moment due to the helicopter lift is the lift times the distance from the lift point (the end of the tower) to the hinge. (Let's assume length, that we're lifting from the end of the tower.)
Now: [weight x (length/2) = lift x length] is the equation for equilibrium. So the helicopter can lift a tower that weighs twice as much as it can carry if the center of gravity of the tower is at the middle. And that's the answer for A, B, and C as well.
That's the answer if you're dealing with high school physics or maybe freshman statics.
In the real world . . .
