The standard Gibbs free energy change is related to the standard enthalpy and entropy changes by the equation $\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$. For $\Delta G^{\circ}$ to decrease sharply with an increase in temperature ($T$), the entropy change of the reaction ($\Delta S^{\circ}$) must be positive. The sign of $\Delta S^{\circ}$ can be predicted by looking at the change in the number of moles of gaseous species ($\Delta n_g$). Let's analyze the given reactions:

1. $\text{C(graphite)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CO(g)}$: Here, $\Delta n_g = 1 - 0.5 = +0.5$. Since $\Delta n_g > 0$, the entropy increases ($\Delta S^{\circ} > 0$).
2. $\text{CO(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)}$: Here, $\Delta n_g = 1 - 1.5 = -0.5$. Since $\Delta n_g < 0$, the entropy decreases ($\Delta S^{\circ} < 0$).
3. $\text{Mg(s)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{MgO(s)}$: Here, $\Delta n_g = 0 - 0.5 = -0.5$. Since $\Delta n_g < 0$, the entropy decreases ($\Delta S^{\circ} < 0$).
4. $\frac{1}{2}\text{C(graphite)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \frac{1}{2}\text{CO}_2\text{(g)}$: Here, $\Delta n_g = 0.5 - 0.5 = 0$. Since $\Delta n_g = 0$, the entropy change is almost zero ($\Delta S^{\circ} \approx 0$). Thus, only in the first reaction is $\Delta S^{\circ}$ positive, causing $\Delta G^{\circ}$ to decrease as the temperature increases.