(a)(i) Let the weekly phone counts be \(A\sim\operatorname{Po}(3.2)\) and \(B\sim\operatorname{Po}(2.9)\). Independence gives \(A+B\sim\operatorname{Po}(6.1)\), so

\[P(A+B=5)=e^{-6.1}\frac{6.1^5}{5!}=0.15785981\ldots=\boxed{0.158}\quad\text{(3 s.f.)}.\]

(a)(ii) Either \((A,B)=(5,0)\) or \((A,B)=(0,5)\). These events are mutually exclusive, so add their probabilities:

\[\begin{aligned}P&=e^{-3.2}\frac{3.2^5}{5!}e^{-2.9}+e^{-3.2}e^{-2.9}\frac{2.9^5}{5!}\\&=e^{-6.1}\frac{3.2^5+2.9^5}{5!}=0.01010516\ldots=\boxed{0.0101}\quad\text{(3 s.f.)}.\end{aligned}\]

(b)(i) The total count in 6 weeks is \(T\sim\operatorname{Po}(6\times6.1)=\operatorname{Po}(36.6)\). Approximate this by \(Z\sim\mathrm N(36.6,36.6)\). More than 30 means at least 31, so apply continuity-corrected boundary \(30.5\):

\[P(T>30)\approx P(Z>30.5)=1-\Phi\left(\frac{30.5-36.6}{\sqrt{36.6}}\right)=0.8433445\ldots=\boxed{0.843}\quad\text{(3 s.f.)}.\]

(b)(ii) The Poisson mean is \(36.6>15\), so the normal approximation is appropriate.

(c) Substitute the two Poisson probabilities: \[e^{-\lambda}=2\mu e^{-\mu}.\]

Take natural logarithms: \[-\lambda=\ln(2\mu)-\mu.\]

Hence \[\boxed{\lambda=\mu-\ln(2\mu)}.\]