m = 3; N = 1000
$
P(X=r) = \frac{e^{-m}e^{r}}{r!} \\
P(X=r) = \frac{e^{-3}3^r}{r!}
$
(i) $
P(X=0) = \frac{e^{-3}3^0}{0!} = e^{-3} = 0.049787
$
Therefore, number of drivers with no accident in a year = 1000 x 0.049787 = 49.787 = 50 drivers.
(ii) $
P(X\gt3) = P(X \geq 4) \\
= 1 - [P(X=0) + P(X=1) + P(X=2) + P(X=3)] \\
= 1 - [\frac{e^{-3}3^0}{0!} + \frac{e^{-3}3^1}{1!} + \frac{e^{-3}3^2}{2!} + \frac{e^{-3}3^3}{3!}] \\
= 1 - [0.049787 + 0.14936 + 0.22404 + 0.22404] \\
= 0.35276938
$
Expected number of drivers with more than 3 accidents = 1000 x 0.35276938 = 352.76938 = 353 drivers