Two vessels A and B of equal volume V_0 are joined by a narrow tube...

Initial state of the system:
Both vessels $A$ and $B$ have volume $V_0$, pressure $p_0$, and temperature $T_0$.
Number of moles in each vessel: $n_A = n_B = \dfrac{p_0 V_0}{R T_0}$.
Total number of moles: $n = n_A + n_B = \dfrac{2 p_0 V_0}{R T_0}$.

When the valve is closed and the pistons are slowly pulled out to double the volumes:
For vessel $A$ (diathermic walls): It remains in thermal equilibrium with the atmosphere. Thus, the process is isothermal.
Final volume $V_A' = 2V_0$
Final temperature $T_A' = T_0$

For vessel $B$ (adiabatic walls): The process is reversible and adiabatic.
Final volume $V_B' = 2V_0$
Using the adiabatic relation $T V^{\gamma - 1} = \text{constant}$:
$T_B' (2V_0)^{\gamma - 1} = T_0 V_0^{\gamma - 1} \Rightarrow T_B' = T_0 2^{1 - \gamma}$

When the valve is opened for a sufficient time:
The gases in vessels $A$ and $B$ mix. Since the walls of vessel $A$ are diathermic, the system continues to exchange heat with the atmosphere. Given "sufficient time", the entire gas mixture will eventually reach thermal equilibrium with the atmosphere.
Thus, the final common temperature of the gas is $T_f = T_0$.

The total final volume of the system is:
$V_f = V_A' + V_B' = 2V_0 + 2V_0 = 4V_0$

Using the ideal gas law for the entire system to find the final pressure $p_f$:
$p_f = \dfrac{n R T_f}{V_f} = \dfrac{\left(\dfrac{2 p_0 V_0}{R T_0}\right) R T_0}{4V_0} = \dfrac{2 p_0 V_0}{4V_0} = \dfrac{p_0}{2}$

Therefore, the new values of the temperature and pressure are $T_0$ and $\dfrac{p_0}{2}$ respectively.

Answer: $T_0$, $\dfrac{p_0}{2}$