Consider the following compounds : K O_2, H_2 O _2 and H_2 S O_4...

In $\text{KO}_2$, potassium is an alkali metal, so its oxidation state is always $+1$.

In $\text{H}_2\text{O}_2$, oxygen is present as a peroxide ion ($\text{O}_2^{2-}$), so the oxidation state of oxygen is $-1$.

In $\text{H}_2\text{SO}_4$, let the oxidation state of sulfur be $x$.

$2(+1) + x + 4(-2) = 0$

$x - 6 = 0$

$x = +6$

The oxidation states of the underlined elements are $+1, -1$ and $+6$ respectively.

Answer: $+1, -1$ and $+6$