To begin with here's a diagram of the initial state of the cylinder:

For the duration of time when there is still sliding taking place then the friction force is at maximum and therefore the acceleration is a constant and so is the torque on the cylinder. This means the following equations of motion hold:

$\displaystyle \omega = \omega_0 + \ddot{\theta}t $ and $\displaystyle v = a t $

where $\displaystyle v$ and $\displaystyle \omega$ are the velocity and angular velocity at the point when sliding stops respectively. Also

$\displaystyle v = r \omega$.

Hence, $\displaystyle v$ can be found in terms of $\displaystyle \ddot{\theta}$ and $\displaystyle a$ as

$\displaystyle v = \frac{a}{a-r \ddot{\theta}} r \omega_0$ .

This equation for the velocity can then be used with the equation of motion relating velocity to distance to find $\displaystyle d_1$, i.e.

$\displaystyle v^2 = u^2 +2aS$ so then

$\displaystyle \color[rgb]{0,0,1}d_1 = \frac{a}{(a-r\ddot{\theta})^2} \frac{r^2 \omega_{0}^{2}}{2}$.

Now all you need to do is find $\displaystyle \ddot{\theta}$ and $\displaystyle a$ and substitute in the above equation. These two can now be found by resolving forces in the diagram.

Friction force is given by

$\displaystyle F_\mu = \mu R$

Resolving perpendicular to the plane we have:

$\displaystyle R = m g \cos \alpha$.

Resolving along the plane

$\displaystyle F_\mu - mg \sin \alpha = m a$ .

From these three equations we find that

$\displaystyle \color[rgb]{0,0,1} a = (\mu \cos \alpha - \sin \alpha) g$.

Finally the torque is given by:

$\displaystyle I \ddot{\theta} = -r F_\mu$ where $\displaystyle \color[rgb]{0,0,1}I = \frac{1}{2} m r^2$.

Hence

$\displaystyle \color[rgb]{0,0,1} \ddot{\theta} = - \frac{2 \mu g \cos \alpha}{r}$ .

So that gives you $\displaystyle d_1$ easily enough!

For the second part you can just use the conservation of energy to work out the further distance travelled, $\displaystyle d_2$, before stopping. So equating kinetic energy lost to the potential energy gained gives:

$\displaystyle \color[rgb]{0,0,1}mg \, d_2 \, \sin \alpha = \frac{1}{2} I \omega^2 +\frac{1}{2} m v^2$

then

$\displaystyle d_{max} = d_1 +d_2$ .