$NaHC{O}_3$ is a salt of weak acid $\left(H_{2}C{O}_3\right)$nd a strong base $(NaOH)$. It undergoes hydrolysis in water as follow:
$$NaHC{O}_{3\left(a_q\right)}\to N{a}_{\left(a_p\right)}^{+}+HC{O}_{3(\text{ ap })}^{-}$$
$$\frac{H_2{O}_{cr}\to H_{(a+)}^{+}+O{H}_{\text{as }}^{-}}{NaHC{O}_{3(k)}+H_{2}O\to N{a}_{(a+)}^{-}+H_2{O}_{(n)}+C{O}_{2(s)}+O{H}_{(a)}^{-}}$$
$O{H}_{(\dot{\alpha })}^{-}$makes the solution alkaline $NaHS{O}_4$ is a salt of strong acid $\left(H_{2}S{O}_4\right)$ and a strong base $(NaOH)$ which does not hydrolyse in water. And since the salt is an acidic one, its solution is acidic.
$NaCl$ is a neutral salt of strong acid $(HCl)$ and strong base $(NaOH)$ and as a result does not undergo hydrolysis in water. So, its solution is neutral.
Note: Only salt of weak acid vs strong base \& strong acid vs weak base undergo hydrolysis in water to give an acidic or neutral solution.