The left branch has a 12 V battery and a 3 Ω resistor.

Step 1. The loops. $I_1 = 4 - 2I_3$ and $I_2 = 2 - 3I_3$, in amperes.

Step 2. The junction. $(4 - 2I_3) + (2 - 3I_3) = I_3$, so $6 = 6I_3$ and $I_3 = 1\ \text{A}$. Then $I_1 = 2\ \text{A}$ and $I_2 = -1\ \text{A}$.

Step 3. (b) $I_2$ is negative: $1\ \text{A}$ flows down through the $4\ \text{V}$ battery, into its positive terminal. It is being charged.

Step 4. (c) Given: $12\ \text{V} \times 2\ \text{A} + 4\ \text{V} \times (-1\ \text{A}) = 20\ \text{W}$. Used: $(2\ \text{A})^2 \times 3\ \Omega + (1\ \text{A})^2 \times 2\ \Omega + (1\ \text{A})^2 \times 6\ \Omega = 20\ \text{W}$.