The variation of equilibrium constant with temperature is given below:
Temperature: ,        Equilibrium Constant: 
$$T_1 = 25^\circ\text{C}$$              $$K_1 = 10$$
$$T_2 = 100^\circ\text{C}$$,           $$K_2 = 100$$
The values of $$\Delta H^\circ$$, $$\Delta G^\circ$$ at $$T_1$$ and $$\Delta G^\circ$$ at $$T_2$$ (in $$\text{kJ mol}^{-1}$$) respectively, are close to [use $$R = 8.314\,\text{JK}^{-1}\text{mol}^{-1}$$]

We are given the equilibrium constant at two temperatures:

$$T_1 = 25°\text{C} = 298\,\text{K}$$, $$K_1 = 10$$

$$T_2 = 100°\text{C} = 373\,\text{K}$$, $$K_2 = 100$$

We need to find $$\Delta H°$$, $$\Delta G°$$ at $$T_1$$, and $$\Delta G°$$ at $$T_2$$.

We use the Van't Hoff equation, which relates the equilibrium constant to temperature:

$$\ln\left(\frac{K_2}{K_1}\right) = \frac{\Delta H°}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$

Substituting the known values:

$$\ln\left(\frac{100}{10}\right) = \frac{\Delta H°}{8.314}\left(\frac{1}{298} - \frac{1}{373}\right)$$

$$\ln(10) = \frac{\Delta H°}{8.314} \times \frac{373 - 298}{298 \times 373}$$

$$2.303 = \frac{\Delta H°}{8.314} \times \frac{75}{111154}$$

$$2.303 = \frac{\Delta H°}{8.314} \times 6.748 \times 10^{-4}$$

Solving for $$\Delta H°$$:

$$\Delta H° = \frac{2.303 \times 8.314}{6.748 \times 10^{-4}} = \frac{19.147}{6.748 \times 10^{-4}} = 28374\,\text{J/mol} \approx 28.4\,\text{kJ/mol}$$

Now we find $$\Delta G°$$ at each temperature using the relation $$\Delta G° = -RT\ln K$$.

At $$T_1 = 298\,\text{K}$$:

$$\Delta G°(T_1) = -8.314 \times 298 \times \ln(10) = -8.314 \times 298 \times 2.303 = -5705\,\text{J/mol} \approx -5.71\,\text{kJ/mol}$$

At $$T_2 = 373\,\text{K}$$:

$$\Delta G°(T_2) = -8.314 \times 373 \times \ln(100) = -8.314 \times 373 \times 4.606 = -14286\,\text{J/mol} \approx -14.29\,\text{kJ/mol}$$

Therefore, $$\Delta H° \approx 28.4\,\text{kJ/mol}$$, $$\Delta G°(T_1) \approx -5.71\,\text{kJ/mol}$$, and $$\Delta G°(T_2) \approx -14.29\,\text{kJ/mol}$$.

The correct answer is Option C: $$28.4,\,-5.71$$ and $$-14.29$$.