The half reactions are
a. `CIO_(3)^(ө)rarrCl^(ө)`
Balancing them separately:
`ClO_(3)^(ө)rarrCl^(ө)`
`ClO_(3)^(ө)+6H^(o+)rarrCl^(ө)+3H_(2)O`
`ClO_(3)^(ө)+6e^(-)+6H^(o+)rarrCl^(ө)+3H_(2)O`
b. `I_(2)rarrIO_(3)^(ө)`
`I_(2)+6H_(2)Orarr21O_(3)^(ө)+12H^(o+)`
`I_(2)+6H_(2)Orarr2IO_(3)^(ө)+10e^(-)+12H^(o+)`
Balance electrons in two half reactions, adding we get:
`5ClO_(3)^(ө)+3I_(2)+6H_(2)OrarrCl^(ө)+6IO_(3)^(ө)+6H^(o+)`