It seems that the OP left, but I liked the question so I will answer it. The notation $_{fg}$ has been addressed in the comments, so I will answer to your "how do I proceed from here?".
Disclaimer I will use Smith, Van Ness, and Abbott book for the water tables, but it doesn't matter for what we are asked. I think they have the same reference as your book, so answers will be almost the same.
Main Idea If $\ce{H2O}$ is a pure gas or liquid, it has two degrees of freedom, and its thermodynamic state is specified two intensive variables. If saturated, then we need only one.
I will talk about states $1A$ (Tank A initially), $1B$ (Tank B initially), and $2$ (final state).
Initial Calculations
$\ce{Tank 1 A}$: We have that $P_{1A} = 400 \; \pu{kPa}$ and the quality $x_{1A} = 0.8$, so it is saturated water, and the thermodynamic state is specified. They give us the quality $x$ because a general thermodynamic property for a pure fluid in LV equilibrium $M$ is obtained by doing $M = x M_{g} + (1 - x)M_l$.
\begin{align}
v_{1A} &= 0.3699928 \; \pu{m^3/kg} \\
u_{1A} &= 2163.0074 \; \pu{kJ/kg}
\end{align}
Of interest will be the initial mass of Tank A
$$ m_{A} = \dfrac{V_A}{v_{1A}} = 0.54055 \; \pu{kg} $$
$\ce{Tank 1 B}$: We have that $P_{2A} = 200 \; \pu{kPa}$ and $T_{2A} = 250 \; \ce{°C}$, thus the thermodynamic state is specified
\begin{align}
v_{2A} &= 1.1989 \; \pu{m^3/kg} \\
u_{2A} &= 2731.4 \; \pu{kJ/kg}
\end{align}
Of interest will be the initial mass of Tank B
$$ m_{B} = \dfrac{V_B}{v_{1B}} = 0.41705 \; \pu{kg} $$
The specific internal energy of state $1$ is
$$ u_1 = \dfrac{m_A u_{1A} + m_Bu_{2A}}{m_A + m_B} = 2410.551 \; \dfrac{\pu{kJ}}{\pu{kg}} $$
Balances
For a closed system, where no work is done, the 1st law states
$$ m\Delta u = Q \rightarrow Q = m(u_2 - u_1) \tag{1} $$
where $m = m_A + m_B$.
Also, since matter does not enter or leave the system, our whole mass resides in the tank volumes. Therefore, the final specific volume is
$$ v_2 = \dfrac{V_A + V_B}{m_A + m_B} \rightarrow
v_2 = 0.73099 \; \dfrac{\pu{m^3}}{\pu{kg}} \tag{2} $$
Final State
With Eq. (2) we are ready to go, since we have our two intensive variables $T_2 = 25 \; \ce{°C}$ and $v_2 = \; 0.73099 \; \pu{m^3/kg}$. However, upon looking at the water tables, this specific volume is between the saturated liquid and saturated gas at $T_2$. Thus, the pressure is informed immediately
$$ \boxed{P_2 = 3.166 \; \pu{kPa}}$$
And the quality and specific internal energy of this saturated water is
\begin{align}
x_2 &= \dfrac{v_2 - v_{2l}(T_2)}{v_{2l}(T_2) - v_{2g}(T_2)} \rightarrow
x_2 = 0.01682 \\
u_2 &= x_2 u_{2g}(T_2) + (1 - x_2)u_{2l}(T_2) = 143.5728 \; \dfrac{\pu{kJ}}{\pu{kg}}
\end{align}
And Eq. (1) yields
$$ \boxed{Q = -2170.858\; \pu{kJ}} $$
The system has given this energy to the surroundings. Makes sense, because we started with two tanks at high $P$ and $T$, and ended at ambient temeperature.
