The first three are simple. Since the expansion is adiabatic, q = 0. Since the external pressure is constant, w = -P_exDV = 1atm x 6 liters = 6 atm liters = -606 J. The change in internal energy is just the sum of these: DU = w + q = -606 J. As for the other two:

For an ideal gas, DH = C_PDT under all conditions.