= Solution
Under <quadratic loss>, the <Bayes estimator under squared error loss> of the latent mean $\lambda$ is its <posterior mean>. For the <Poisson-uniform posterior mean>, the <prior distribution> has density $1/2$ on $(1,3)$ and the one-count <Poisson distribution> likelihood is $\lambda e^{-\lambda}$. Consequently the <Bayesian posterior> density is proportional to $\lambda e^{-\lambda}$ on that interval, with normalizing integral $\int_1^3\lambda e^{-\lambda}\,d\lambda$.
By conditional independence, $\mathbb E(X_2\mid\lambda,X_1)=\lambda$, so the <law of total expectation> makes the posterior predictive <expected value> equal to this same <posterior mean>. It is
$$
\mathbb E(\lambda\mid X_1=1)=\frac{\int_1^3\lambda^2e^{-\lambda}\,d\lambda}{\int_1^3\lambda e^{-\lambda}\,d\lambda}.
$$
The required antiderivatives are $-(\lambda+1)e^{-\lambda}$ and $-(\lambda^2+2\lambda+2)e^{-\lambda}$. Evaluating at the two endpoints gives
$$
\boxed{\widehat m_{\rm Bayes}=\frac{5e^{-1}-17e^{-3}}{2e^{-1}-4e^{-3}}
=\frac{5e^2-17}{2e^2-4}\approx1.85054.}
$$
This estimate differs from $13/7$: the <Bühlmann credibility estimate> is an optimal affine rule, while the <Bayes estimator under squared error loss> optimizes over all rules and uses the truncated uniform prior through its exact <Bayesian posterior>.
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