You can for all practical purposes "ignore" solids and pure liquids in calculations. This is because the activity or concentration of a solid or pure liquid is constant and thus taken to be "1". Therefore they don't affect calculation of the equilibrium constant.

Therefore in the second set of problems, the denominator of the equilibrium constant expression is 1, and we basically ignore it and just write out the numerator. That's why it looks as if we totally ignored the solids and pure liquids in the solubility equilibrium.

If you reversed the reaction so that the aqueous ions were on the left and the solid was on the right, then your numerator would be 1 and the denominator would be the product of the ion concentrations.