Answer
$$0.100 \space M \space HCl : Neutral$$ $$0.100 \space M \space HF : Basic$$
Work Step by Step
$$HCl(aq) + KOH(aq) \longrightarrow KCl(aq) + H_2O(l)$$
At the equivalence point, the reactants are completely consumed. Thus, the solution only has $KCl$ and $H_2O$. Since these two are neither acidic or basic, the pH is neutral.
$$HF(aq) + KOH(aq) \longrightarrow KF(aq) + H_2O(l)$$
$F^-$ is the conjugate pair of $HF$, which makes it a weak base. Thus, the solution at the equivalence point is basic.
