Help! I only got ONE HOUR! Physcis problem

A rescue helicopter lifts a 80 kg person straight up by means of a cable. The person has an upward acceleration of 0.70 m/s2 and is lifted from rest through a distance of 13 m.

(a) What is the tension in the cable?

(b) How much work is done by the tension in the cable?

(c) How much work is done by the person's weight?

(d) Use the work-energy theorem and find the final speed of the person.

A) The tension in the cable is enough to hold the person up, plus the amount to accelerate the person:
T=mg+ma
T=m(g+a)
T=80*(9.8+0.70)

B)Take that force from part a (the tension, T) and multiply it by the distance traveled:
W=T*13

C)Take the person's weight (as a force) and multiply it by the distance traveled:
W=m*g*13

D)The work energy theorem tells us that the net work equals the change in kinetic energy. The work from part B is going up and the work from part C is going down. Take the answer from B and subtract the answer from C:
Wnet=T*13-m*g*13
Wnet=(T-m*g)*13

That is equal to Kf-Ki. It says "starts from rest" which means that Ki=0. Therefore Wnet is equal to Kf. We also know that:
Kf=(1/2)m*v^2
where v is the velocity at the end that you want. Solving for v we get:
v=Sqrt(2Kf/m)
But remember that Kf=Wnet. Plug in Wnet and calculate!