Firstly, according to the stability constant table of Kf and its complexes, [Zn(NH3)6]2+ is a very stable complex. Therefore, in the presence of NH3, Zn2+ exists almost completely in the form of [Zn(NH3)6]2+.

This means that we need to calculate the concentration of [Zn(NH3)6]2+ in 0.6 L of 1.2 M NH3, from which we can infer the concentration of Zn2+. Then, we can use Ksp to determine the amount of dissolved Zn(OH)2.

Because [Zn(NH3)6]2+ is very stable, its concentration is equal to the total Zn2+ concentration, that is:

[Zn(NH3)6]2+ = [Zn2+]

Therefore, [Zn2+] =1.2m.

Now consider [Zn(OH)2]. Let [Zn(OH)2] = x M, then:

zn2+][oh-]^2 =( 1.2 m)(2x)^2 = 4* 10^-3

Through simple algebraic operation, we can get:

X = [OH-] = 0.029

Therefore, [[Zn (OH) 2] = 2x = 0.058mol/L/L.

Therefore, in 0.6 L of 1.2 M NH3, Zn(OH)2 can dissolve at most 0.058 mol.