Nice problem in rotation

till the time there is no pure rolling,
force acting is μmg
acceleration is μg
and torque is μmgR

angular acceleration is α = μmgR/(2/5mR²) = 5μg/2R
v_t=-v₀ + μgt
ω(t)=5μg/2Rt

time to reach velocity 0 wrt ground => t=v₀/μg

ω(t)=5μg/2Rt

so velocity of the bottom most point is given by:
-v₀ + μgt + 5μg/2t = v₁

when t=v₀/μg

-v₀ + μgv₀/μg + 5μg/(2v₀/μg) = v₁

5/2v₀ = v₁

v₀ = 2/5v₁

part 2 is also solved above..

part 3 is easy by v²=u²+2as!