finding cation of unknown salt in solution from precipitate
Among the options for the second salt in solution, one is less likely to be found in the presence of $\ce{K_2SO_4}$.

These solutions for analysis are prepared by adding decent amounts of two salts with good solubility to water. Under these conditions, a double displacement reaction may occur and in one case, a precipitate will be formed:

$\ce{K_2SO_4 + CaCl_2 -&gt; 2KCl + CaSO_4}$

I doubt that the person preparing the solutions for analysis used calcium chloride, since 

this would have required to remove the precipitate by filtration or decanting off the supernatant solution
the concentration of $\ce{Ca^{2+}}$ in the solution would be rather low. Remember that these qualitative tests are not necessary suitable for trace analysis.

The second salt therefore is most likely sodium chloride ($\ce{NaCl}$), potassium chloride ($\ce{KCl}$) or magnesium chloride ($\ce{MgCl_2}$).

I would do a flame test first. If you see a yellow coloured flame, the second cation is the sodium cation ($\ce{Na+}$). Similarly if you see a lilac coloured flame, the second cation is the potassium cation ($\ce{K+})$ and if you initially see a colourless flame and finally see an intense white coloured flame,the second cation is the magnesium cation ($\ce{Mg^{2+}}$).
If you want to test for the magnesium cation ($\ce{Mg^{2+}}$), place a drop of the solution on a spot plate, make sure it's alkaline by adding aqueous $\ce{NaOH}$ if necessary, and add a bit of a titan yellow solution. This azo dye with its misleading name (there is no titanium in it) will form a red precipitate with the magnesium cation ($\ce{Mg^{2+}}$).