Thermodynamic transformation Why is it that any reversible thermodynamic transformation is quasi-static? Also, why is the converse not necessarily true?
 A: The most important reason to limit the scope to quasi-static transformations is, because you are doing equilibrium thermodynamics and therefore always (maybe implicitly) assume that the system is in an equilibrium state.
The question itself appears to me like you are thinking of the difference between adiabatic (no exchange of heat) and isentropic (no change in entropy). An isentropic process is always adiabatic (by virtue of Clausius' inequality). However, the reverse is not true. Take for example the Gay-Lussac experiment (Gas is in one half and then free to expand into the other half, assuming perfect isolation there is no transfer of heat, but the entropy increases)
A: A thermodynamic process is called reversible if an infinitesimal change of the external conditions is capable of reversing the process. 
A thermodynamic process is called quasi-static if it is a dense succession of equilibrium states. Roughly speaking, a quasi-static process is one that can be represented by a continuous curve in the thermodynamic configuration space.
If the process is reversible then we can continually change the external conditions and make the system evolve along a dense succession of equilibrium states. It is quasi-static.
On the other hand a quasi-static process can involve friction or non reversible exchange of heat i.e. heat exchanged trough a finite difference of temperature. In this case an infinitesimal change of the external conditions is no longer able to invert the direction of evolution of the system. It is irreversible.
