How does the rate of reaction change with an increase in temperature?

For an endothermic reaction $X \rightarrow Y$,the activation energies for the forward and backward reactions are $E_f$ and $E_b$ respectively. Then,in general:

$A$ reaction rate constant is given by $k = 1.2 \times 10^{14} \, e^{-25000/RT} \, sec^{-1}$. It means

Decomposition of a hydrocarbon follows the equation $k = (5.5 \times 10^{11} \text{ s}^{-1}) e^{\frac{-28000 \text{ K}}{T}}$. The activation energy of the reaction is . . . . . . $\text{kJ mol}^{-1}$. (Nearest Integer) Given: $R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$

For the reaction $A + B \rightarrow C + D$,$\Delta H = -20 \ kJ \ mol^{-1}$. The activation energy for the forward reaction is $85 \ kJ \ mol^{-1}$. The activation energy for the reverse reaction is ......... $kJ \ mol^{-1}$.

An exothermic reaction $X \rightarrow Y$ has an activation energy of $30 \ kJ \ mol^{-1}$. If the energy change $\Delta E$ during the reaction is $-20 \ kJ \ mol^{-1}$,then the activation energy for the reverse reaction in $kJ \ mol^{-1}$ is $...$.