The key to this is to understand that in half of all cases where two aces were drawn, the second random step will pick the heart 1/2 the time.

Thus - the first step selects 5 out of 6 possible outcomes. In the second step, 4 of the 5 cases have fixed outcomes; the other (Aa) has two posssible outcomes.

Thus there are six possible final states:

1 a D showing a (1/5) 2 a d showing a (1/5) 3 A D showing A (1/5) 4 A d showing A (1/5) 6 A a showing a (1/10) 6 A a showing A (1/10)

The answer to the second question is yes in cases 2, 3, and 6. The collective probability of these cases is 1/5 + 1/5 + 1/10 = 1/2.

1/10 over 1/2 is 1/5.