A and C are very simple. Since there is no acceleration in both cases, the weight of the passenger is just [tex]mg=(70kg)(9.8\frac{m}{s^2})=686 N[/tex] (about 690 N or 150 lbs with correct sig-figs). For B, we must first find the acceleration of the elevator. Since [tex]v=v_{0}+at [/tex], we can write that [tex]11\frac{m}{s}=0\frac{m}{s}+a(4.5s)[/tex]. Solving for a, we get that [tex]a\approx2.44\frac{m}{s^2}[/tex]. Plugging this into Newton's second law, we get that [tex]F=ma=(70kg)(2.44\frac{m}{s^2})\approx171 N[/tex]. When calculating the apparent weight, this gets added onto the 686 N from A and C, so the final apparent weight is about 857 N, or 860 N with the correct number of sig-figs.
tl;dr:
A - 690 N
B - 860 N
C - 690 N
