Let us consider an endothermic reaction which is non-spontaneous at the freezing point of water. However, the reaction…

🧠 Endothermic + non-spontaneous at low T but spontaneous at high T → both $\Delta H > 0$ and $\Delta S > 0$ For $\Delta G = \Delta H - T\Delta S$ to flip from positive to negative as $T$ increases, the entropy term must grow large enough to overcome $\Delta H$. This requires $\Delta S > 0$.

Since "endothermic" is given: $\Delta H > 0$. Spontaneous at $373\,\text{K}$ but not at $273\,\text{K}$: $T_e = \Delta H / \Delta S$ lies between 273 and 373 K. This requires $\Delta S > 0$ so that at high enough $T$, $T\Delta S > \Delta H$.

$\boxed{\text{Answer: (a) Both } \Delta H \text{ and } \Delta S \text{ are positive}}$