Answer from cs61c-ax (Amanda Alfonso 15906918) for Question 3
Since the equation simplifies to E = A(BnotC + CnotB), you need 3 AND gates and 1 OR gate.  The first AND gate produces BnotC and the second CnotB.  These are then passed through an OR gate to produce (BnotC + CnotB).  Finally that is passed through the third AND gate with A to produce E = ABnot(C) + ACnot(B).  Thus, there are 8 wires needed: 2 to each of the gates' inputs (and 9 wires if you include the output).