Answer
The tension in the bottom part of the rope is 380 N.
The tension in the top part of the rope is 760 N.
Work Step by Step
The tension in the bottom part of the rope $F_b$ provides the force to hold the third climber. This tension force $F_b$ must be equal to the component of the third climber's weight which is directed down the hill ($mg ~sin(\theta)$):
$F_b = mg ~sin(\theta)$
$F_b = (75 ~kg)(9.80 ~m/s^2) ~sin(31.0^{\circ})$
$F_b = 380 ~N$
The tension in the bottom part of the rope is 380 N.
The tension in the top part of the rope $F_t$ provides the force to hold both the second and third climbers. This tension force $F_t$ must be equal to the component of the second and third climber's weight which is directed down the hill ($2mg ~sin(\theta)$):
$F_t = 2mg ~sin(\theta)$
$F_t = (2)(75 ~kg)(9.80 ~m/s^2) ~sin(31.0^{\circ})$
$F_t = 760 ~N$
The tension in the top part of the rope is 760 N.
