These are all pretty straightforward
a) use the Poisson distribution $P[k] = \dfrac{\lambda^k e^{-\lambda}}{k!},~k=0,1,\dots$
here $\lambda = 0.5~flaws/pane$
$P[\text{at least 1 flaw}] = 1 - P[\text{0 flaws}]$
b) let $X$ be the number of flaws, distributed as in (a)
$E[Profit] = 5 P[X=0] - 3 P[X>0]$
c) here $\lambda = 2~flaws/pane$
find $P[X=0]$