Answer
The power of battery 2 is $~~-0.158~W$
Work Step by Step
Let $i_1$ be the current through $R_1$ and let's assume the direction of the current is to the right.
Let $i_2$ be the current through $R_2$ and let's assume the direction of the current is to the left.
Let $i_3$ be the current through $R_3$ and let's assume the direction of the current is down.
Using the junction rule:
$i_3 = i_1+i_2$
Using the loop rule counterclockwise around the right loop:
$1.00-2.00~i_2-5.00~i_3 = 0$
$1.00-2.00~i_2-5.00~(i_1+i_2) = 0$
$1.00-7.00~i_2-5.00~i_1 = 0$
$-5.00+35.0~i_2+25.0~i_1 = 0$
Using the loop rule clockwise around the left loop:
$3.00-4.00~i_1-5.00~i_3 = 0$
$3.00-4.00~i_1-5.00~(i_1+i_2) = 0$
$3.00-9.00~i_1-5.00~i_2 = 0$
$21.0-63.0~i_1-35.0~i_2 = 0$
We can add these two equations to find $i_1$:
$16.0-38.0~i_1 = 0$
$i_1 = \frac{16.0}{38.0}$
$i_1 = 0.421~A$
The current through resistance 1 is $~~0.421~A$
We can find $i_2$:
$1.00-7.00~i_2-5.00~i_1 = 0$
$1.00-7.00~i_2-(5.00)~(0.421) = 0$
$7.00~i_2 = -1.105$
$i_2 = \frac{-1.105}{7.00}$
$i_2 = -0.158~A$
The negative sign indicates that the current flows through $R_2$ to the right.
The current in battery 2 is $0.158~A$ and the current goes in the opposite direction to the battery's alignment in the circuit.
We can find the power of battery 2:
$P = i~\mathscr{E}_2$
$P = (0.158~A)(-1.00~V)$
$P = -0.158~W$
The power of battery 2 is $~~-0.158~W$
