For adding up equations, you can't directly add $E^0$,unless you know that the no. of electronic transfres are same in both the reactions given.So, it is safe to calculate by $\Delta$$G^0$ calculation,
For $HOCl$ reaction , you have,
$HOCl + 2e^- + H^+ = Cl^- + H_2O$ -- $\Delta$$G^0$= -$nFE^0$ = $-2*F*1.481$
By inverting the First reaction, sign of $E^0$ will be changed and thus,
$Cl^- + 3H_2O = ClO_3^- + 6e^- + 6H^+$ --- $\Delta$$G^0$=-$nFE^0$ = $-6*F*(-1.446)$
Adding the two reaction ,
$HOCl + 2H_2O = ClO_3^- + 4e^- + 5H^+$ -- $\Delta$$G^0$= - $(4*F*E^o)$
$\Delta$$G^0$ will be added and for the whole reaction,
$E^0(total) $ =
$\Delta$$G^0$(total)/$nF$= $-F*(2*1.481-6*1.446)/(-4*F)$)=$-1.4285$ $V$.
So, $E^0$ for the oxidation from $HOCl$ to $ClO_3^-$ is $-1.4285$ $ V$.
Thus, the answer is different from simply adding the $E^0$ for reactions which is not correct for all the cases(is correct when electron transfer is same for both reactions).