a) $p=P[\text{at least 5 calls}] = 1 - P[\text{4 calls or less}] = 1 - \sum \limits_{k=0}^4 \dfrac{(3)^k e^{-3}}{k!} = 0.184737$
b) using $p$ from part (a) this is just a binomial distribution $n=7, ~p$
$P[\text{at least 1 cell gets 5 calls}] = 1 - P[\text{0 cells get 5 calls}] = 1 - (1-p)^7 = 0.760622$