#### Answer

(a) Zach's apparent weight is 784 N
(b) Zach's apparent weight is 1050 N

#### Work Step by Step

Let $F_N$ be the normal force of the elevator pushing up on Zach. Note that this force $F_N$ will be equal to Zach's apparent weight.
(a) When the elevator is descending at a constant speed, Zach's acceleration is zero. Then $F_N$ will be equal in magnitude to mg.
$F_N = mg$
$F_N = (80~kg)(9.80~m/s^2)$
$F_N = 784~N$
Zach's apparent weight is 784 N.
(b) We can find the acceleration directed upward when the elevator is braking.
$a = \frac{v_f-v_0}{t}$
$a = \frac{0-10~m/s}{3.0~s}$
$a = -3.3~m/s$
The magnitude of the acceleration is $3.3~m/s^2$ and it is directed upward.
We can use the acceleration to find $F_N$.
$\sum F = ma$
$F_N - mg = ma$
$F_N = m(g+a)$
$F_N = (80~kg)(9.80~m/s^2+3.3~m/s^2)$
$F_N = 1050~N$
Zach's apparent weight is 1050 N.