physics elevator problem?

Question

A person stands on a scale in an elevator.
The maximum and minimum scale readings
are 652:4 N and 301:8 N, respectively.
The acceleration of gravity is 9:8 m=s2 :
Assume the magnitude of the acceleration
is the same during starting and stopping, and
determine the weight of the person. Answer
in units of N.

Determine the person's mass. Answer in units
of kg.

Determine the acceleration of the elevator.
Answer in units of m=s2.

Answer 1

And the acceleration of the elevator is e.

m*g = 477.1 N
m*(g+e) = m*g + m*e = 652.4 N

Solve for e.

Answer 2

If the magnitude of acceleration of the elevator at both stopping and starting is the same, it will exert the same pseudoforce at each time, but in opposite directions. So on starting to move up (or stopping moving down) the force on the person's feet will be their own weight (W), and the force of the floor pushing up (F). On stopping moving up (or starting to move down), the force on the person's feet will be their own weight, less the pseudoforce of the floor pulling away.
So

W+F = 652.4N

and

W-F = 301.8N

The weight is the average of the two figures. The mass is this figure divided by 9.8 ms-2.