Answer
$Al^{3+}$ presents, $Na^{+}$ possibly presents, $Mg^{2+}$ and $Ag^{2+}$ definitely absent.
Work Step by Step
When our solution is treated with hydrochloric acid, $Cl^{-}$ ions are supplied. Because precipitation does not happen, we can deduce that $Ag^{+}$ ions are definitely absent, as $AgCl$ solubility is almost non-existent.
Our solution is then treated with $H_{2}S$ twice, once at $pH=1$ and the other at $pH=8$. At $pH=1$, there is no precipitation and we cannot deduce anything from it, as $Al^{3+}$, $Na^{+}$ and $Mg^{2+}$ don't precipitate at this situation. However, at $pH=8$, $[OH^{-}]$ is profoundly higher, causing $Al(OH)_3$ to precipitate. Because $Al^{3+}$ is the only ion capable of precipitation in this case, we definitely know that $Al^{3+}$ exists.
Finally, the solution is treated with $(NH_4)_2HPO_4$, and there is no precipitation. Thus, we know that $Mg^{2+}$ does not exist, as $MgNH_4PO_4$ should have precipitated.
Throughout this whole test, $Na^{+}$, being an alkali ion, could not be identified using methods based on precipitation, since it remains soluble in all situations mentioned above. Hence, we will mark it as "possibly present."
