Assigning current direction in mesh analysis

No, they don't have to be in the same direction. You just need to remember which direction you assumed because your final answer will tell you if the current is in the same direction as your assumed direction. If you solve for current and you get a positive answer, it means your assumed current is in the same direction as the actual current. If you solve for current and you get a negative answer, it means the actual current is in the opposite direction from what you assumed.

I'm not sure what is meant by "passive configuration", but the current supplied by the voltage source is opposite the direction of \$i_s\$. This is because

In a single source circuit, the direction of positive current flows out of positive terminal of a voltage source. However, in a multiple source circuit, this condition may not be valid.

You are using superposition, so for the analysis of \$i_1\$ you have a single source circuit. Since the current from the voltage source is in the opposite direction of \$i_s\$, \$i_s\$ is negative.

As for the case where the voltage source is turned off, the current through the (upper) \$5\Omega\$ resistor is zero because it is short-circuited by the voltage source (\$v_1\$ and \$v_2\$ are the same node). You can view these two paths as a current divider with the short-circuited path as a \$0\Omega\$ resistor -- if the current entering the two paths is \$i\$, then the current through the short-circuited path is

$$\frac{5\Omega}{5\Omega + 0\Omega}i = i$$

Alternatively, the current through the \$5\Omega\$ path is

$$\frac{0\Omega}{5\Omega + 0\Omega}i = 0$$

In other words, all the current flows through the short circuited path and none of it through the \$5\Omega\$ resistor. This means you can ignore the upper \$5\Omega\$ resistor as it has no effect on the circuit (no current passes through it, nor is there any voltage across it). Without this \$5\Omega\$ resistor, you should see that there are only two resistors and the desired current is easily solved with a current divider as you have stated.