You'd need "24.6 mL" of oxygen for the combustion of "CO" and "10.0 mL" of oxygen for the combustion of hydrogen.
Let's start with the combustion of carbon monoxide. The balanced chemical equation is
2CO_((g)) + O_(2(g))-> 2CO_(2(g))
Since no mention of pressure or temperature is made, I'll assume we're at STP, where pressure is equal to 1 atm and temperature is equal to 273.15 K. Moreover, at STP, 1 mole of any ideal gas occupies 22.4 L, or "22.4 dm"^3. Since you have "50 cm"^3 of carbon monoxide, the number of moles you have is
n = V/V_("molar") = (50 * 10^(-3)L)/("22.4 L/mol") = 0.0022 "moles"
Notice that you have a 2:1 mole ratio between "CO" and "O"_2, which means that the number of oxygen moles you have is
"0.0022 moles CO" * ("1 mole O"_2)/("2 moles CO") = 0.0011 "moles"
This translates into a volume of
V = n_("oxygen") * V_("molar") = "0.0011 moles" * 22.4 "L"/"mole"
V = "0.0246 L" = 24.6 "mL"
This is the balanced chemical equation for the combustion of hydrogen
2H_(2(g)) + O_(2(g)) -> 2H_2O_((g))
The same strategy applies for this reaction as well. The number of hydrogen moles is
n = V/V_("molar") = (20*10^(-3)"L")/("22.4 L/mol") = 0.00089 "moles"
Once again, we have a 1:1 mole ratio to apply, so the number of oxygen moles is
"0.00089 moles H"_2 * ("1 mole O"_2)/("2 moles H"_2) = 0.00045 "moles"
Therefore, the volume of oxygen needed is
V = n_("oxygen") * V_("molar") = "0.00045 moles" * 22.4 "L"/"mole"
V = "0.010 L" = 10.0 "mL"