Answer
If the values of $\mu_1$ and $\mu_2$ are reduced, then the value of $h$ is also reduced.
Work Step by Step
In part (a), we found $F$, the least horizontal pull by the hands and push by the feet that will keep the climber stable.
We can consider the torque about a rotation axis at the climber's center of mass.
The two friction forces at the hands and feet are directed upward, and there is a counterclockwise torque from these friction forces.
The rock walls exert a force $F$ on the hands and the feet, and there is a clockwise torque from these forces.
Let $h_1$ be the vertical distance of the feet below the center of mass.
Let $h_2$ be the vertical distance of the hands above the center of mass.
Note that $h_1+h_2 = h$
We can find an expression for $h$:
$\tau_{net} = 0$
$(d)(F~\mu_1)+(d+w)(F~\mu_2)-(h_1)(F)-(h_2)(F) = 0$
$(h_1)(F)+(h_2)(F) = (d)(F~\mu_1)+(d+w)(F~\mu_2)$
$(h_1)+(h_2) = (d)~(\mu_1)+(d+w)~(\mu_2)$
$h = (d)~(\mu_1)+(d+w)~(\mu_2)$
If the values of $\mu_1$ and $\mu_2$ are reduced, then the value of $h$ is also reduced.
