Why are reactions with higher activation energy more sensitive to temperature?
The first thing to note is that the rate constant with a large activation barrier will be far smaller than one with a small barrier at a given temperature. If the barrier is very small (e.g. 5 kJ/mol) at, say, room temperature, almost all collisions between the reactive molecules have sufficient energy to react and the barrier is crossed and this leads to a fast reaction. This means that increasing the temperature cannot lead to much of an increase, because the reaction is going almost as fast as it can at the lower temperature. 

Now consider the case with a large barrier. In this case the rate constant is far smaller than when the barrier is small, and very few collisions have enough energy  to cross the barrier. Now increase the temperature and the number of filled energy levels the molecule has increased greatly, exponentially so by the Boltzmann distribution.  Now very many more collisions have sufficient energy to react and so the rate constant increases greatly, although it may still be small in value. 

If you plot the Arrhenius equation (given in your text) vs temperature the slope of this gives the rate of change with temperature. With larger activation energies the slope gets larger. This is the effect also described in your text. 

The figure shows rate constants calculated with the activation energies $E_a$ shown (in kJ/mol). You can see how rapidly the rate constant increases with temperature when the activation energy is large, i.e. the slope is greater when the activation energy is larger, than when the activation energy is small. (The rate constant $\displaystyle k=A_0e^{-E_a/(RT))}$ are with $A_0 =10,\; (E_a=10);\; A_0= 10^3,\;(E_a=20)$ and $A_0=10^6,\;(E_a=40)$ with $A_0$ chosen only to get them on the same scale). 

The gradient of the rate constant is $\displaystyle \frac{dk}{dT}=E_a\frac{A_0e^{-E_a/(RT)}}{RT^2}$ which shows that the slope depends on the activation energy.