For Q1, the Keplerian orbit in the sentence you refer to is a hypothetical, not a consequence.   *If* it loses angular momentum and *if* it remains in a Keplerian orbit *then* the orbit must have a smaller radius.  This follows from the equation.

Q2.  Note the the slide doesn't say that A slows down and B speeds up.  It says that friction *tries* to slow down A and *tries* to speed up B.  But if this process is done slowly and continuously, that doesn't happen.  As the energy and angular momentum is pulled from A, the orbit lowers in such a way that the speed increases.  At no time does A actually slow down.

Q3.  I'd prefer the term "added" or "gained", but yes.  Any angular momentum lost by A is gained by B.  The sum remains constant.

Q4.  Correct.

U1.  Why do you ask about $v_\phi$ being constant?  It's not.  As the ring contracts, it speeds up.  As it enlarges, it slows down.


> I have difficulties with "tries to slow down A and tries to speed up B" and saying nothing of both occur : So how the lose or win of angular momentum can be acheived ?

Because speed isn't directly related to the angular momentum.  There is an interaction (*viscous friction* in the slide), and this might initially cause a slight slowdown in the particles.  But the orbit of the particles changes as well and ends up speeding them up (by more than the friction slowed them down).

The inner ring loses angular momentum and gains orbital speed at the same time.

>You indicated that Q4 was correct : can we say the same thing for inward motion of mass : if ring A loses angular momentum, there will be an inward motion of mass from ring A to a lower orbit ? 

Yes.  We're assuming the ring has a particular mass that is roughly constant.  So the only change in $\sqrt{GMr}$ is the radius (which is decreasing).