Answer
The direction of $i_2$ is down.
Work Step by Step
The equivalent resistance of $R_7$ and $R_8$ is $R_{78} = 4.0~\Omega$
The equivalent resistance of $R_1, R_2$ and $R_{78}$ is $R_{1278} = 2.0~\Omega+2.0~\Omega+4.0~\Omega = 8.0~\Omega$
The equivalent resistance of $R_6$ and $R_{1278}$ is $R_{12678} = 4.0~\Omega$
The equivalent resistance of $R_3$ and $R_{12678}$ is $R_{123678} = 4.0~\Omega+4.0~\Omega = 8.0~\Omega$
The equivalent resistance of $R_4$ and $R_5$ is $R_{45} = 3.0~\Omega+1.0~\Omega = 4.0~\Omega$
Let $i_4$ be the current through resistance 3 and let's assume the direction of the current is to the left.
Using the loop rule counterclockwise around the left loop using $R_{123678} = 8.0~\Omega$:
$16-8.0~i_4 = 0$
$8.0~i_4 = 16$
$i_4 = 2.0~A$
The potential difference across $R_3$ is $(2.0~A)(4.0~\Omega) = 8.0~V$
Then the potential difference across $R_6$ is $8.0~V$
The current through $R_6$ is $\frac{8.0~V}{8.0~\Omega} = 1.0~A$
Then the current through $R_{1278}$ is $~~2.0~A-1.0~A = 1.0~A$
Since $R_7=R_8$, then half of this current passes through $R_8$
Therefore, $i_2 = 0.50~A$
Since the current in the left loop flows from top to bottom, the direction of $i_2$ is down.
