Answer
The least horizontal pull by the hands and push by the feet that will keep the climber stable is $~~337~N$
Work Step by Step
In order for the net horizontal forces on the climber to be zero, the hands and feet must exert the same magnitude of force $F$ on the rock.
Then the total friction force exerted upward on the climber must be equal in magnitude to the climber's weight.
We can find the minimum possible force $F$:
$F~\mu_1+F~\mu_2 = mg$
$F = \frac{mg}{\mu_1+\mu_2}$
$F = \frac{(55~kg)(9.8~m/s^2)}{0.40+1.2}$
$F = 337~N$
The least horizontal pull by the hands and push by the feet that will keep the climber stable is $~~337~N$.
