The exact volumes of 1 M NaOH solution required to neutralise 50 mL of 1 M $$H_3PO_3$$ solution and 100 mL of 2 M $$H_3PO_2$$ solution, respectively, are:

First, consider $$H_3PO_3$$ (phosphorous acid). This is a diprotic acid with the structure $$HP(O)(OH)_2$$. It has two ionizable $$O-H$$ groups and one non-ionizable $$P-H$$ bond. So 1 mole of $$H_3PO_3$$ reacts with 2 moles of NaOH.

For 50 mL of 1 M $$H_3PO_3$$: moles of $$H_3PO_3 = 0.05$$ mol. Moles of NaOH required $$= 2 \times 0.05 = 0.1$$ mol. Volume of 1 M NaOH $$= \frac{0.1}{1} = 0.1$$ L $$= 100$$ mL.

Next, consider $$H_3PO_2$$ (hypophosphorous acid). This is a monoprotic acid with the structure $$H_2P(O)(OH)$$. It has only one ionizable $$O-H$$ group and two non-ionizable $$P-H$$ bonds. So 1 mole of $$H_3PO_2$$ reacts with 1 mole of NaOH.

For 100 mL of 2 M $$H_3PO_2$$: moles of $$H_3PO_2 = 0.2$$ mol. Moles of NaOH required $$= 1 \times 0.2 = 0.2$$ mol. Volume of 1 M NaOH $$= \frac{0.2}{1} = 0.2$$ L $$= 200$$ mL.

The volumes required are 100 mL and 200 mL, which corresponds to option (3).