(d) > 500N for time Δt1 and Δt3 and 500N for Δt2.
Explanation:
During the constant velocity for time Δt2 there is no relative movement between the man and the seat and only force applied by the seat on the man is Normal Force which is equal to weight of the man =500 N.
But during the acceleration (Δt1) and deceleration (Δt3) the seat exerts an addition force of friction on man in horizontal direction. Let this be F.
Now the force by seat on man is resultant of 500 N and F,
=√{(500)²+F²} N, which is >500 N.