To answer these questions, we first need to understand two key concepts:

Normality (N), which differs from molarity and relates to the number of equivalents per liter of solution.
The equivalent weight of the compound, which in this case involves understanding how many electrons are transferred in the half-cell reaction.

Part (i): Let's calculate based on the half-cell reaction:

$$BrO_3^- + 6H^+ + 6e^- \rightarrow Br^- + 3H_2O$$

This reaction shows each mole of $$BrO_3^-$$ consumes 6 equivalents (6 moles of electrons). The molar mass of sodium bromate ($$NaBrO_3$$) is $$22.99 + 79.904 + 48.00 = 150.894\, g/mol$$.

To find the weight of sodium bromate needed:

1. The equivalent weight of $$NaBrO_3$$ = $$\frac{Molar\, mass}{6} = \frac{150.894}{6} = 25.149\, g/equiv$$.

2. To make a 0.672 N solution in 85.5 mL, first convert mL to L: $$85.5\, mL = 0.0855\, L$$.

3. Calculate the total equivalents needed: $$0.672\, eq/L \times 0.0855\, L = 0.05748\, equivalents$$.

4. Calculate the mass of $$NaBrO_3$$ required: $$0.05748\, equivalents \times 25.149\, g/equiv = 1.446\, g$$.

To find the molarity (M) of the solution:

The molarity would be the number of moles of $$NaBrO_3$$ in the given volume of solution:

1. Number of moles = $$\frac{1.446\, g}{150.894\, g/mol} = 0.00958\, moles$$.

2. Molarity = $$\frac{0.00958\, moles}{0.0855\, L} = 0.112\, M$$.

Part (ii): Consider the revised half-cell reaction:

$$2BrO_3^- + 12H^+ + 10e^- \rightarrow Br_2 + 6H_2O$$

Here, 2 moles of $$BrO_3^-$$ are reduced by 10 electrons. Each mole of $$BrO_3^-$$ consumes 5 equivalents (5 electrons):

1. The equivalent weight of $$NaBrO_3$$ in this case = $$\frac{Molar\, mass}{5} = \frac{150.894}{5} = 30.179\, g/equiv$$.

Calculating the mass required to produce a 0.672 N solution:

1. Total equivalents required: $$0.672\, eq/L \times 0.0855\, L = 0.05748\, equivalents$$

2. Mass of $$NaBrO_3$$ required: $$0.05748\, equivalents \times 30.179\, g/equiv = 1.734\, g$$.

Calculating the molarity based on these equivalents:

1. Number of moles = $$\frac{1.734\, g}{150.894\, g/mol} = 0.0115\, moles$$.

2. Molarity = $$\frac{0.0115\, moles}{0.0855\, L} = 0.134\, M$$.

In summary, the weight and molarity needed under each half-cell reaction are calculated by considering the number of electrons transferred per mole of reactant and adjusting the equivalent weight accordingly.