Answer
The climber must exert a minimum normal force of 490 N.
Work Step by Step
We can assume that the force of static friction pushing up on the climber is equal in magnitude to the climber's weight $mg$:
$F_N (0.80) + F_N (0.60) = mg$
$F_N (1.40) = (70.0 ~kg)(9.80 ~m/s^2)$
$F_N = \frac{(70.0 ~kg)(9.80 ~m/s^2)}{1.40} = 490 ~N$
Therefore, the climber must exert a minimum normal force of 490 N.