Ans: (a)( log n)/pie (b) [(M/m-n)/(M/m+n)]g
Here n is eta
Hint: The ratio of tension in such cases is given by \(e^{\mu\theta}\) where \(\theta\) is the upper angle between the radii of the pully joining the centre of the pulley to the extremities of the part of the string in contact with the pulley and in this specific case, \(\theta=\pi\).
To prove the above result, one can take small length of the string in contact with the pulley and consider tensions as T & T+dT, normal reaction dN and friction \(df=\mu dN\), and then apply the tangential and normal equilibrium conditions of forces for the small length of the string and integrate from, say T1 to T2.
For further help, you may refer to I.E. Irodov's Problems in General Physics page 284. If you still have problem with this, you may send me a private message in virtual classroom.