The oxidation number of an atom in a molecule or an ion is essentially a theoretical charge that the atom would have if all the bonds to atoms of different elements were completely ionic. To determine the oxidation numbers of sulfur in the given compounds: S₈, S₂F₂, and H₂S, let's analyze them one by one.

1. In S₈: This is an elemental form of sulfur, comprising of sulfur atoms only. In elemental forms, the atoms are not bonded to atoms of a different element, but only to themselves. In such cases, the oxidation number is always 0. Hence, the oxidation number of sulfur in S₈ is 0.

2. In S₂F₂: This is a molecule consisting of sulfur and fluorine. Fluorine is more electronegative than sulfur and virtually always has an oxidation number of -1 in its compounds. Let's denote the oxidation number of sulfur as $x$. Since there are two fluorine atoms, each contributing -1, and the overall molecule is neutral, we get:

$2x + (2 \times -1) = 0$

$2x - 2 = 0$

$2x = 2$

$x = +1$

Therefore, the oxidation number of sulfur in S₂F₂ is +1.

3. In H₂S: In this compound, hydrogen has an oxidation number of +1 (except when bonded to metals where it can be -1). Since there are two hydrogens, their total contribution is +2. Let's denote the oxidation number of sulfur in this compound as $x$. The compound is neutral overall, so:

$2(+1) + x = 0$

$2 + x = 0$

$x = -2$

Therefore, the oxidation number of sulfur in H₂S is -2.

Comparing with the given options, the correct answers are 0, +1, and -2 for S₈, S₂F₂, and H₂S, respectively. Therefore, the correct option is Option A: 0, +1, and -2.