How much faster would a reaction proceed at $25\,^{\circ}C$ than at $0\,^{\circ}C$ if the activation energy is $65\,kJ/mol$?

For the following reactions:
$A \xrightarrow{700 \ K}$ Product
$A \xrightarrow[\text{catalyst}]{500 \ K}$ Product
it was found that $E_{a}$ is decreased by $30 \ kJ/mol$ in the presence of a catalyst. If the rate remains unchanged,the activation energy for the catalysed reaction is (Assume pre-exponential factor is same):

At $298 \text{ K}$ temperature,the activation energy for the reaction $x_2 + y_2 \rightarrow 2xy + 20 \text{ kJ}$ is $15 \text{ kJ}$. What will be the activation energy for the reaction $2xy \rightarrow x_2 + y_2$?

For an exothermic reaction $A \to B$,the activation energy is $65 \ kJ \ mol^{-1}$ and the heat of reaction is $-42 \ kJ \ mol^{-1}$. The activation energy for the reverse reaction $B \to A$ would be....... $kJ \ mol^{-1}$

What names apply to chemical species corresponding to locations $1$ and $2$ on this reaction coordinate diagram?
Location $1$ $-$ Location $2$

The rate constant of a reaction at $25\,^oC$ is $1 \times 10^{-3}\,s^{-1}$. If the rate of the reaction doubles when the temperature is increased to $35\,^oC$,the activation energy of the reaction is .......... $kJ\, mol^{-1}$.