Q2 is the successor of Q1 under input {a}, and Q1 = ({s0,s2},{s0}). Since the second component {s0} of Q1 is not empty, we have to compute the first component of Q2 as sucσ∃(Q) and the second component as sucσ∃(Qf)∩F. State s0 has no transition for input {a}, and state s2 has a transition to s1, so that sucσ∃({s0,s2}) = {s1}. For the second component, we already checked that s0 has no transition for input {a}, so sucσ∃({s0}) = {}, and the intersection with the accepting states F does not change this.

So, note that the second component is sucσ∃(Qf)∩F and not sucσ∃(Q)∩F in case that Qf is nonempty.