- Using your first approach, van't Hoff :

$$
\begin{eqnarray} 
&& \ln \frac{k_2}{k_1} = -\frac{∆H^\circ}{R} (\frac{1}{T_2} - \frac{1}{T_1})
\end{eqnarray} 
$$

For an exothermic reaction $\Delta H <0$

Now , Let  $T_2>T_1$   therefore   $K_2<K_1$ i.e. equilibrium constant decreases.

- Using your second approach, 

$$
\begin{eqnarray}
\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ  
\end{eqnarray}
$$
$$
\begin{eqnarray}
\Delta G = \Delta H - T \Delta S 
\end{eqnarray}
$$ 
$$
\begin{eqnarray}
\Delta G = \Delta G^\circ + RT \ln(Q)
\end{eqnarray}
$$ 
At equilibrium $ \Delta G = 0 $ and $Q= K_{eq}$  and  hence
$$ \begin{eqnarray} 
&& K=e^{\frac{-\Delta G^\circ}{RT}} =e^{\frac{-(\Delta H^\circ - T \Delta S^\circ)}{RT}} 
\end{eqnarray} $$

For Temperature $T_1$ and Equilibrium Constant $K_1$
$$
\begin{eqnarray}
K_1=e^{\frac{-(\Delta H^\circ - T_1 \Delta S^\circ)}{RT_1}} 
\end{eqnarray}
$$
Similarly, For Temperature $T_1$ and Equilibrium Constant $K_1$
$$
\begin{eqnarray}
K_2=e^{\frac{-(\Delta H^\circ - T_2 \Delta S^\circ)}{RT_2}} 
\end{eqnarray}
$$

Dividing the above two equations, you get 
$$
\begin{eqnarray} 
&& \frac{k_2}{k_1} = e^{-\frac{∆H^\circ}{R} (\frac{1}{T_2} - \frac{1}{T_1})}
\end{eqnarray} 
$$

The Van't Hoff Equation !!! And it would give the same result as before.

I didn't conclude Dependence of $K_{eq}$ and T from $\Delta G^\circ$ because if a reaction is exothermic, it only tells that $\Delta H^\circ <0$ and doesn't tell anything about $\Delta G^\circ$ .