Why is the equilibrium constant increasing in this exothermic reaction?
An exothermic reaction has a reduced equilibrium constant at higher $T$ because while the contribution of the change in the entropy of the system is a fixed quantity (for a small $T$ change), the effect of transferring heat to the surroundings is reduced at higher $T$ (because it causes a smaller change in the entropy of the surroundings). Mathematically this is described by the van't Hoff equation:
$$\left(\frac{\partial \ln K_\textrm{eq}}{\partial T}\right)_p = \frac{\Delta H^\circ}{RT^2}$$
On the other hand the change in $\Delta G$ with temperature is given by $\Delta S$:
$$\left(\frac{\partial\Delta G}{\partial T}\right)_p = -\Delta S$$
Since $\Delta G =\Delta G^\circ +RT\ln Q = -RT\ln (K/Q)$ one can write
$$\left(\frac{\partial\Delta G}{\partial T}\right)_p = \left(\frac{\partial\Delta G^\circ}{\partial T}\right)_p +R\ln Q\\ = -\Delta S^\circ+R\ln Q$$
Therefore if you want to see how $ K_\textrm{eq}$ depends on $T$ you should focus on $\Delta H^\circ$. If you want to see how $\Delta G$ depends on $T$ you should inspect $-\Delta S^\circ+R\ln Q$.
It is somehow unfortunate that the temperature dependence of $\Delta G$ and the equilibrium constant are different, because it leads to endless mix-ups, but it reminds you that they describe different things.