S1 = nu x. ([]x & mu y. (b | b | [] y)) is alternation free since least fixpoint mu y. (b | b | [] y) nested in the greatest fixpoint does not depend on the fixpoint variable of the outer fixpoint. Hence, you can first compute the state of the inner fixpoint with one fixpoint iteration, and with these states, you can compute the fixpoint of the outer fixpoint also with a single fixpoint iteration.

S2 = nu x. (<:>x & nu y. ((a|x) & []y)) is also alternation free since the two nested fixpoints are of the same kind, so that there is no alternation. Alternation means that least fixpoints are nested in greatest fixpoints or vice versa.