Differential amplifier, superposition theorem, output derivation, what is a loop here?
Since V1 is grounded, the upper loop gives you the equation

$$ I_{top}R_1 = -I_{top}R_2 $$

which implies that the current in the upper loop is zero. Thus, the voltage at the positive input terminal  of the op-amp is 0 volts. 

Therefore, the current on the bottom loop is

$$I_{bottom} = \frac{V_2 - 0}{R_3} = \frac{V_2}{R_3} $$

so by using by Ohm's law with resistor R4,

$$ \frac{0 - V_{out2}}{R_4} = \frac{-V_{out2}}{R_4} = I_{bottom}$$

so

$$V_{out2} = -\frac{R_4V_2}{R_3}$$