Which of the following oxidation states are the most characteristic for lead and tin respectively?

All the group 14 elements show tetravalency, which can be explained by the shifting of one electron from $ns$ to the vacant $np$ orbital; the four orbitals then give four $sp^3$ hybrid orbitals.
When the $ns^2$ electrons of the outermost shell do not participate in bonding, they are called an inert pair, and the effect is called the inert pair effect.
The last three elements of the group, Ge, Sn and Pb, have a tendency to form divalent as well as tetravalent ions.
The inert pair effect increases from Ge to Pb, so the stability of the tetravalent state decreases and that of the divalent state increases down the group.
That is why $Pb^{2+}$ is more stable than $Pb^{4+}$ (Pb(IV) compounds are strong oxidising agents), while $Sn^{4+}$ is more stable than $Sn^{2+}$ (Sn(II) compounds are reducing agents).
So the most characteristic oxidation states are +2 for lead and +4 for tin.