Adiabatic expansion of an ideal gas
In the following answer I am using the Physics convention of work done. Where, $W= \int PdV$ and the first law of thermodynamics is as follows: $q = \Delta U + W$. If you want answers in the chemistry conventions than just reverse the sign of the work done. Internal energy change and heat change will be same in both conventions.
As you said you have done the first part, I will continue with the second.
First of all, $C_V=C_P - R = 8.58- 2 = 6.58$
As the process is adiabatic,
$$q=0\\
\Delta U=-W\\
nC_V(T_2-T_1)=-nRP_{ext} (\frac{T_2}{P_2} - \frac{T_1}{P_1}) \\
T_2= 247.67 K = -25. 53 ^\circ C\\\\$$
$\\Now, \\\\$
$$W= P_{ext} \Delta V\\
W= nRP_{ext} (\frac{T_2}{P_2} - \frac{T_1}{P_1}) \\
W= 599.2 cal\\\\$$
$$\Delta U= -W \\
\Delta U= -599. 2 cal\\\\$$
$$\Delta H= \Delta U + \Delta (PV) \\
\Delta H= \Delta U + \Delta (nRT)\\ 
\Delta H= \Delta U + nR\Delta T\\
\Delta H= (-599. 2 -181. 32) cal\\
\Delta H= -780. 52 cal\\\\$$