= Solution
For $N\sim\operatorname{Pois}(\lambda)$, the <Poisson distribution> gives
$$
H(N)=\lambda\log_2e-\lambda\log_2\lambda+\mathbb E\log_2(N!).
$$
This <information entropy> is finite: $\log_2(n!)\leq n^2$ for integer $n\geq0$, while $\mathbb E N^2=\lambda+\lambda^2$.
Fix $0<\lambda_1<\lambda_2$. Take <independent random variables> $U\sim\operatorname{Pois}(\lambda_1)$ and $V\sim\operatorname{Pois}(\lambda_2-\lambda_1)$. Their <convolution> is
$$
\mathbb P(U+V=n)
=e^{-\lambda_2}\sum_{j=0}^n
\frac{\lambda_1^j(\lambda_2-\lambda_1)^{n-j}}{j!(n-j)!}
=e^{-\lambda_2}\frac{\lambda_2^n}{n!},
$$
by the <binomial theorem>. Thus $U+V$ has <Poisson distribution> with parameter $\lambda_2$. Conditioning on $V=v$ only translates $U$, so $H(U+V\mid V)=H(U)=F(\lambda_1)$. The <entropy monotonicity under independent addition> yields $F(\lambda_2)\geq F(\lambda_1)$.
In fact the inequality is strict. The <covariance> $\operatorname{Cov}(U+V,V)=\operatorname{Var}(V)=\lambda_2-\lambda_1$ is positive, so $U+V$ and $V$ are not <independent random variables>. Their <information entropies> are finite, and the equality criterion from part (a) excludes equality. Therefore the <Poisson entropy monotonicity> gives the stronger conclusion
$$
\boxed{\lambda_2>\lambda_1>0\ \Longrightarrow\ F(\lambda_2)>F(\lambda_1).}
$$
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