It is b ∧ nu x. <>x, not b^… (^ is xor, ∧ is and).
I'm not sure whether I understand your notes correctly but you already wrote y = {s0, s4}. First, we do nu y. <>y. y0 = all states, y1=all states except s7 because it as no successor hence, also none to y0, y2=y1. Now, we intersect it with the states satisfying b (s0, s4, s7). As we already kicked out s7, only s0, s4 are left.
To solve the outer mu, we just iteratively add predecessors that satisfy a. The iteration likes likes: s0,s4, then s0, s4, s5, then s0, s4, s5, then s0, s4, s5, s1, then s0, s4, s5, s1, s2, then fixpoint.