300 mL of 0.50 M HClO + 400 mL of 0.50m NaClO. How do I setup an ICE table for this? What is its net ionic reaction?

There is no net ionic reaction between ##”HClO”## and ##”ClO”^-##.
This ##color(red)(“is”)## your buffer solution.
The buffer components are the weak acid ##”HClO”## and its conjugate base ##”ClO”^-##.
The only net ionic reaction that you need is the equation for the ionization of ##”HClO”##:
##”HClO” + “H”_ 2″O” ⇌ “H”_ 3″O”^+ + “ClO”^(-)##; ##”p”K_”a” = 7.53##
You still have to calculate the moles of each component.
##”Moles of HClO” = 0.300 cancel(“L”) × “0.50 mol”/(1 cancel(“L”)) = “0.15 mol”##
##”Moles of ClO”^(-) = 0.400 cancel(“L”) × “0.50 mol”/(1 cancel(“L”)) = “0.20 mol”##
The Henderson-Hasselbalch Equation is;
##”pH” = “p”K_”a” + log((“[ClO”^(-)”]”)/(“[HClO]”))##
##”pH” = 7.53 + log((0.20 cancel(“mol”))/(0.15 cancel(“mol”))) = 7.53 +0.12 = 7.67##