#### Answer

(a) The person's apparent weight is $567.3~N$
(b) The person's apparent weight is $628.3~N$

#### Work Step by Step

(a) We can find the person's mass:
$mg = 598~N$
$m = \frac{598~N}{9.80~m/s^2}$
$m = ~61.0~kg$
Let $F_N$ be the upward force exerted on the person by the elevator. Note that $F_N$ is the person's apparent weight. The elevator is accelerating downward. We can find the person's apparent weight:
$\sum F = ma$
$mg-F_N = ma$
$F_N = m~(g-a)$
$F_N = (61.0~kg)(9.80~m/s^2-0.50~m/s^2)$
$F_N = 567.3~N$
The person's apparent weight is $567.3~N$
(b) Let $F_N$ be the upward force exerted on the person by the elevator. Note that $F_N$ is the person's apparent weight. The elevator is accelerating upward. We can find the person's apparent weight:
$\sum F = ma$
$F_N-mg = ma$
$F_N = m~(g+a)$
$F_N = (61.0~kg)(9.80~m/s^2+0.50~m/s^2)$
$F_N = 628.3~N$
The person's apparent weight is $628.3~N$