In the preparation of$\mathrm{HBr}$ or $\mathrm{HI}, \mathrm{NaX} (\mathrm{X} : \mathrm{Br}, \mathrm{I})$ is treated with ${\mathrm{H}}_3{\mathrm{PO}}_4$ and not by concentrated ${\mathrm{H}}_2{\mathrm{SO}}_4$ since

$\mathrm{HBr}$ or HI cannot be prepared by heating ${\mathrm{Br}}^{\ominus }$ or ${\mathrm{I}}^{\ominus }$ with conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$ because $\mathrm{HBr}$ and HI are strong reducing agents and reduces 

${\mathrm{H}}_2{\mathrm{SO}}_4$ to ${\mathrm{SO}}_2$ and ${\mathrm{Br}}^{\ominus }$ or ${\mathrm{I}}^{\ominus }$ are oxidized to ${\mathrm{Br}}_2$ or ${\mathrm{I}}_2$

respectively. 

$\begin{array}{r}\mathrm{NaX}+{\mathrm{H}}_2{\mathrm{SO}}_4⟶{\mathrm{NaHSO}}_4+\mathrm{HX}\\ {\mathrm{H}}_2{\mathrm{SO}}_4+2\mathrm{HX}⟶{\mathrm{SO}}_2+{\mathrm{X}}_2+2{\mathrm{H}}_2\mathrm{O}\\ (\mathrm{X}=\mathrm{Br}\text{ or }\mathrm{I})\end{array}$

Hence $\mathrm{HBr}$ and HI are prepared by heating  ${\mathrm{Br}}^{\ominus }$ and ${\mathrm{I}}^{\ominus }$ respectively with ${\mathrm{H}}_3{\mathrm{PO}}_4$

$\begin{array}{r}3\mathrm{NaBr}+{\mathrm{H}}_3{\mathrm{PO}}_4⟶{\mathrm{Na}}_3{\mathrm{PO}}_4+3\mathrm{HBr}\\ 3\mathrm{NaI}+{\mathrm{H}}_3{\mathrm{PO}}_4⟶{\mathrm{Na}}_3{\mathrm{PO}}_4+3\mathrm{HI}\end{array}$