Answer
(a) $\frac{5}{7}h$
(b) $h$
(c) Without friction, the marble's rotational energy can not be converted to potential energy, only the regular kinetic energy can. Thus, no extra height gained from the rotational energy in (a).
Work Step by Step
(a) Total energy at the top of the left side (height $h$):
$$ E_1 = U_1 = mgh$$
Total energy at the bottom of the bowl:
$$ E_2 = K_{\mathrm{rotational}} + K_{\mathrm{translational}} = \frac{I \omega^2}{2} + \frac{mv_{cm}^2}{2} = \frac{I \omega^2}{2} + \frac{5I\omega^2}{4} = \frac{7I\omega^2}{4} $$
Total energy when maximum height $x$ is reached on the right side:
$$ E_3 = U_3 + K_{\mathrm{rotational}} = mgx + \frac{I \omega^2}{2} $$
As far as we know, total energy of the system is conserved, so these are all equal. Putting $E_1 = E_2$:
$$ mgh = \frac{7I\omega^2}{4} \Rightarrow \frac{2mgh}{7} = \frac{I\omega^2}{2} $$
Substituting this expression into the above expression for $E_3$ and setting $E_3 = E_1$ yields:
$$mgx + \frac{2mgh}{7} = mgh \Rightarrow x =\frac{5}{7}h $$
(b) With friction on both sides of the bowl, the marble would reach the same height $h$ as it started from, since all its kinetic energy at the bottom would be converted back into potential energy.
(c) Without friction, the marble's rotational energy cannot be converted to potential energy, only the translational kinetic energy can. Thus, no extra height gained from the rotational energy in (a).
