Conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$ cannot be used to prepare $\mathrm{HBr}$ from $\mathrm{NaBr}$ because it:

Using concentrated ${\mathrm{H}}_2{\mathrm{SO}}_4$, HBr cannot be created from $\mathrm{NaBr}$ because $\mathrm{HBr}$ is a strong reducing agent and reduces  ${\mathrm{H}}_2{\mathrm{SO}}_4$ to ${\mathrm{SO}}_2$ and get itself oxidised to  ${\mathrm{Br}}_2$.

$\begin{array}{l}\mathrm{NaBr}+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{NaHSO}}_4+\mathrm{HBr}\\ {\mathrm{H}}_2{\mathrm{SO}}_4+2\mathrm{HBr}\to {\mathrm{SO}}_2+{\mathrm{Br}}_2+2{\mathrm{H}}_2\mathrm{O}\end{array}$