${\mathrm{S}}^{2-}$ and ${\mathrm{SO}}_3^{2-}$ can be distinguished by using :

Both the reagents can be used to distinguish between ${\mathrm{S}}^{2-}$ and ${\mathrm{SO}}_3^{2-}$.

Test with $(C{H}_{3}COO{)}_{2}Pb$

$\begin{array}{l}{\mathrm{S}}^{2-}+{\left({\mathrm{CH}}_3\mathrm{COO}\right)}_2\mathrm{Pb}\to \mathrm{PbS}\downarrow +2{\mathrm{CH}}_3{\mathrm{COO}}^{-}\\ {\mathrm{SO}}_3^{2-}+{\left({\mathrm{CH}}_3\mathrm{COO}\right)}_2\mathrm{Pb}\to {\mathrm{PbSO}}_4\downarrow +2{\mathrm{CH}}_3{\mathrm{COO}}^{-}\end{array}$

Test with ${\mathrm{Na}}_2\left[\mathrm{Fe}(\mathrm{CN}{)}_5\mathrm{NO}\right]:$

$\begin{array}{l}{\mathrm{S}}^{2-}+{\mathrm{Na}}_2\left[\mathrm{Fe}(\mathrm{CN}{)}_5\mathrm{NO}\right]\to {\mathrm{Na}}_4\left[\mathrm{Fe}(\mathrm{CN}{)}_5\mathrm{NOS}\right]\mid \\ {\mathrm{SO}}_3^{2-}+{\mathrm{Na}}_2\left[\mathrm{Fe}(\mathrm{CN}{)}_5\mathrm{NO}\right]\to {\mathrm{Na}}_4\left[\mathrm{Fe}(\mathrm{CN}{)}_5{\mathrm{NO}}_2{\mathrm{SO}}_3\right]\end{array}$

Hence, the correct option is (a) and (b).