The number of calls per hour is the random variable with mean $5$.

The required probability of the call cost that exceeds $\frac{4}{2}$ in an hour

$\Rightarrow P\left(r>2\right)=1-P\left(r\le 2\right)$

$=1-\left(e^{-5}+\frac{e^{-5}5}{1!}+\frac{e^{-5}{5}^2}{2!}\right)$

$=1-e^{-5}\left(1+5+\frac{25}{2}\right)$

$=1-\frac{1}{e^5}\left(\frac{37}{2}\right)$

$\therefore P\left(r>2\right)=\frac{2e^5-37}{2e^5}$