(a) \[\mathrm P(X\ge3)=1-e^{-1.5}\left(1+1.5+\frac{1.5^2}{2!}\right)=0.191.\]

(b) The sum has distribution \(S\sim\operatorname{Po}(4.5)\). Thus \[\mathrm P(3\le S\le5)=e^{-4.5}\left(\frac{4.5^3}{3!}+\frac{4.5^4}{4!}+\frac{4.5^5}{5!}\right)=0.529.\]

(c) For independent values, \(T\sim\operatorname{Po}(1.5n)\), approximated by \(N(1.5n,1.5n)\). Continuity correction gives \[\frac{330.5-1.5n}{\sqrt{1.5n}}=\Phi^{-1}(1-0.0391)\approx1.761.\] Let \(u=\sqrt{1.5n}>0\). The equation becomes \[u^2+1.761u-330.5=0,\qquad u=\frac{-1.761+\sqrt{1.761^2+4(330.5)}}{2}.\] Then \(n=u^2/1.5\approx200\). As \(n\) is a count, the required value is \(200\). Substituting \(n=200\) into the normal approximation reproduces 0.0391 to 3 significant figures.