(a)(i) Sum the three Poisson probabilities: \[P(1\le X\le3)=e^{-2.5}\left(2.5+\frac{2.5^2}{2!}+\frac{2.5^3}{3!}\right)=0.675491\ldots=\boxed{0.675}\quad\text{(3 s.f.)}.\]

(a)(ii) Independence gives \(X+Y\sim\operatorname{Po}(2.5+\lambda)\). The given equality becomes

\[e^{-2.5}\frac{2.5^5}{5!}=e^{-(2.5+\lambda)}\quad\Longrightarrow\quad e^{-\lambda}=\frac{2.5^5}{5!}=0.81380208\ldots.\]

Taking logarithms, \[\lambda=-\ln\left(\frac{2.5^5}{5!}\right)=0.20603808\ldots.\]

Therefore \[P(X+Y=1)=e^{-(2.5+\lambda)}(2.5+\lambda)=0.18076590\ldots=\boxed{0.181}\quad\text{(3 s.f.)}.\]

MS clarification: the bracketed intermediate line in the official MS reverses the signs of the exponents. The correct relation is \(e^{-\lambda}=0.8138\ldots\), as derived above. The MS’s subsequent value of \(\lambda\) and final answer are correct.

(b) The sample size is large and the success probability is small, so a suitable approximation is \(V\sim\operatorname{Po}(np)=\operatorname{Po}(4)\). Then

\[P(W=2)\approx P(V=2)=e^{-4}\frac{4^2}{2!}=0.14652511\ldots.\]

This rounds to \(0.15\) to 2 decimal places and \(0.147\) to 3 decimal places. The given binomial value \(0.146\) also rounds to \(0.15\) to 2 decimal places, so the values agree at 2 decimal places but differ at 3 decimal places, as required.