(a)   H₂O is sp³ hybridized, and oxygen atom has 2 bond pair and 2 lone pair. So the angle between two O $$-$$ H bond is 104.5^o

(b)   In H₂S molecule, central atom S is a 3rd period element and according to Dragos rule, when a 3rd period or higher period element overlap with a small element whose electronegetivity is low then that over lapping cannot be a effective overlapping, because of higher size of 3rd or higher periods element and smaller size of low electronegative element.

Here in H₂S, H atom has very low electronegativity and smaller in size so it create more negative charge around sulphur(s) atom and size of S^$$-$$2 ion increases, because of this higher size difference efficiency overlapping is not possible. So, any hybridization do not happen in between S and H atom. Lone pair electrons of S atom is present in the pure orbital like s, p_x, p_y, p_z and it look like this :



As we know the angle between and p_y is 90^o, so bond angle is 90^o.

(c)   NH₃ has sp³ hybridization and N atom has 3 bond pair and one lone pair so bond angle is 107^o

(d)   SO₂ is sp² hybridized and bond angle is 117^o.