In a S2O3^2- O is always 2-. So we have 3*2-= -6. The overall charge is -2. So this means S2 must give us +4, or each S is +2.
For the products the same thing. O is -2. This means each S is +2.5. This does not seem right? You can't have a fractional oxidation number. This is because the oxidation number calculated this way is an average. Since there are 4 sulfurs this is an average of them.
If you draw the lewis structure for S4O6^(2-) you get O3-S-S-S-S-O3
In organic chemistry you should have learned how to calculate the oxidation states separately instead of using the average way. The sulfur connected to O3 has an oxidation of +5. To get this look at S in the periodic table. It has 6 valance electrons normally. It is bonded to 3 oxygen's, the oxygens are more electronegative and pull the electrons away from S, so we pretend as though it lost the electrons. The s is also bond to another s. Since they are the same element they share equally and each s gets 1 electron from the bond. So after doing all of this you will see that that the S attached to the oxygens only has 1 electron from the s-s bond, but it should have a valence of 6. Its oxidation number is +5. The S on the other side is also +5. The S-S in the middle have an oxidation state of 0. They have 2 lone pairs on each s and form 2 bonds to other s elements. So 2 lone pairs + 2 equal bonds = 6 electrons. Oxidation state is 0. So now we have 5+5+0+0=10 and 10/4 = 2.5 (the average oxidation state).
For B. The reactants ClO-. O is 2-. So Cl has to be +1.
For C. In the reactants O is 2-. S is +6. In the products O is 2- and S is +6.
For D. Zn is 0 in the reactants. Elements are 0. In the products you should have memorized the polyatomic ions and know that SO4 is 2- overall. This gives us s of +6 and o 2-. If SO4^(2-) then Zn has to be 2+.