Answer from cs61c-ed (Cory Benavides 14101530) for Question 3
7 wires:  One wire for each A&B (=w1), A&C (=w2).  One for each !C (=w3), !B (=w4).  One for each w1&w3 (=w5), w2&w4 (=w6).  And finally one wire for w5+w6 (=w7).  We don't count the inputs of the expression as wires.