Both examples are redox reactions.
Let's take the general form of the first one (I'm not going to write the ions for #Ba(OH)_2#, I'll just leave it like this):
#Ba(OH)_(2(aq)) + H_2O_(2(aq)) + 2ClO_(2(aq)) -> Ba(ClO_2)_(2(s)) + 2H_2O_((l)) + O_(2(g))#
If you assign oxidation numbers to all the atoms involved in the reaction, you'll get
#Ba^(+2)(O^(-2)H^(+1))_2 + H_2^(+1)O_2^(-1) + 2Cl^(+4)O_2^(-2) ->Ba^(+2)(Cl^(+3)O_2^(-2))_2 + 2H_2^(+1)O^(-2) + O_2^(0)#
So, #"Cl"#'s oxidation number changes from +4 in #"ClO"_2#, to +3 in #"Ba(ClO"_2)_2#, while #"O"#'s oxidation nubmer changes from -1 in #"H"_2"O"_2#, to 0 in #"O"_2#.
#"Cl"^(+4) + 1"e"^(-) -> "Cl"^(+3)#
#"O"^(-1) -> "O"^(0) + 1"e"^(-)#
For the second reaction, you'll have
#I_2^(+5)O_5^(-2) + 5C^(+2)O^(-2) -> I_2^(0) + 5C^(+4)O_2^(-2)#
Iodine's oxidation number changes from +5 in #"I"_2"O"_5#, to 0 in #"I"_2#, while carbon's oxidation number changes from +2 in #"CO"#, to +4 in #"CO"_2#.
#"C"^(+2) -> "C"^(+4) + 2"e"^(-)#
#"I"^(+5) + 5"e"^(-) -> "I"^(0)#
