Superposition theorem
from where the value of 1 comes?


It's the 1 ohm resistance of r2.  As other answers point out, this step is basic current division.  Using conductances, \$G = \frac{1}{R}\$, current division has the form of voltage division (in fact, current division is the dual of voltage division).

Voltage division for two series connected resistors:

\$V_{R_1} = V_S \dfrac{R_1}{R_1 + R_2}\$

Current division for two parallel connected resistors:

\$I_{R_1} = I_S \dfrac{G_1}{G_1 + G_2} \$

Note that these equations are duals.  If you know one of these, you get the other by duality.

Applying the 2nd equation to your problem:

\$I_2 = I_1 \dfrac{G}{G + g_2} = I_1 \dfrac{\frac{1}{4}}{\frac{1}{4} + \frac{1}{1}} = I_1 \dfrac{1}{1 + 4}\$