A rigid, well-insulated, sealed steel vessel of internal volume 0.30 m³ contains 1.20 kg of wet steam at an initial pressure of 8.5 bar and a dryness fraction of 0.85. The vessel is subsequently cooled until its pressure falls to 2 bar.

At $P_1 = 8.5 \text{ bar}$ and $x_1 = 0.85$, from saturated steam tables:
$v_f = 0.001118 \text{ m}^3/\text{kg}$
$v_g = 0.2246 \text{ m}^3/\text{kg}$

At $P_1 = 8.5 \text{ bar}$ and $x_1 = 0.85$, from saturated steam tables:
$u_f = 735.0 \text{ kJ/kg}$
$u_{fg} = 1849.6 \text{ kJ/kg}$

Given mass $m = 1.20 \text{ kg}$.

The initial specific volume is $\boxed{0.1911 \text{ m}^3/\text{kg}}$, the initial specific internal energy is $\boxed{2307.16 \text{ kJ/kg}}$, and the total internal energy is $\boxed{2768.6 \text{ kJ}}$.

Since the vessel is rigid and sealed, the total volume and mass remain constant. Therefore, the specific volume remains constant.

At $P_2 = 2 \text{ bar}$, from saturated steam tables:
$v_f = 0.001061 \text{ m}^3/\text{kg}$
$v_g = 0.8857 \text{ m}^3/\text{kg}$

At $P_2 = 2 \text{ bar}$, from saturated steam tables:
$u_f = 504.49 \text{ kJ/kg}$
$u_{fg} = 2024.6 \text{ kJ/kg}$

At $P_2 = 2 \text{ bar}$, $v_f = 0.001061 \text{ m}^3/\text{kg}$ and $v_g = 0.8857 \text{ m}^3/\text{kg}$.
Since $v_f < v_2 < v_g$ ($0.001061 < 0.1911 < 0.8857$), the steam is in the wet vapor region.
The specific volume at the final state is $\boxed{0.1911 \text{ m}^3/\text{kg}}$, the final dryness fraction is $\boxed{0.2148}$, and the final specific internal energy is $\boxed{939.58 \text{ kJ/kg}}$. The phase region is wet vapor.

For a closed system with a rigid vessel, work $W=0$. The First Law simplifies to $Q = \Delta U = U_2 - U_1$.

The negative sign indicates that heat is rejected by the steam.
The heat transfer during cooling is $\boxed{-1641.1 \text{ kJ}}$. Heat is rejected by the steam, as indicated by the negative sign, meaning energy is leaving the system.