= Solution
Normalize the specified state by $\langle0|0\rangle=1$ and write $\langle A\rangle=\langle0|A|0\rangle$. Its zero mode is a <coherent state>, so any normally ordered product of zero-mode <creation operators> and <annihilation operators> can be evaluated using $\psi_0\mapsto\sqrt{N_0}$ and $\psi_0^\dagger\mapsto\sqrt{N_0}$. In particular $\langle(\psi_0^\dagger)^2\psi_0^2\rangle=N_0^2$, rather than $N_0(N_0-1)$, which would apply to a zero-mode <number state>.
Sort the <momentum-space contact-interaction Hamiltonian> by its number of zero-mode factors. Four such factors give $uN_0^2/(2V)$. Three zero-mode factors and one nonzero <momentum> violate <momentum conservation>, so no such term survives. With two zero-mode factors there are three possibilities:
* Two zero-mode <creation operators> leave the opposite-<momentum> pair $\psi_{\mathbf k}\psi_{-\mathbf k}$.
* Two zero-mode <annihilation operators> leave $\psi_{\mathbf k}^\dagger\psi_{-\mathbf k}^\dagger$.
* One zero-mode factor on each side leaves $\psi_{\mathbf k}^\dagger\psi_{\mathbf k}$. Choosing which of the two <creation operators> and which of the two <annihilation operators> are zero gives four placements.
The sums over $\mathbf k\ne0$ include both members of each opposite-<momentum> pair, consistently with the overall factor $1/2$. Terms with one zero-mode factor contain three nonzero-mode factors; their <expected values> vanish in the centered <multimode squeezed vacuum>. Indeed each $\psi$ is linear in $\varphi,\varphi^\dagger$, and an odd product has zero <vacuum expectation value>. Terms with no zero-mode factors are independent of $N_0$ at fixed $\theta_k$. Therefore
$$
\boxed{\langle H_{\rm int}\rangle=\frac{uN_0^2}{2V}
+\frac{uN_0}{2V}\sum_{\mathbf k\ne0}
\left[\langle\psi_{\mathbf k}\psi_{-\mathbf k}\rangle
+\langle\psi_{\mathbf k}^\dagger\psi_{-\mathbf k}^\dagger\rangle
+4\langle\psi_{\mathbf k}^\dagger\psi_{\mathbf k}\rangle\right]
+O(N_0^0).}
$$
To evaluate these <vacuum expectation values>, use the inverse <Bogoliubov transformation> and its <Hermitian adjoint>:
$$
\psi_{\mathbf k}=c_k\varphi_{\mathbf k}-s_k\varphi_{-\mathbf k}^\dagger,
\qquad
\psi_{\mathbf k}^\dagger=c_k\varphi_{\mathbf k}^\dagger-s_k\varphi_{-\mathbf k}.
$$
Since $\varphi_{\mathbf k}|0\rangle=0$, the only nonzero two-factor <vacuum expectation value> is $\langle\varphi_{\mathbf k}\varphi_{\mathbf q}^\dagger\rangle=\delta_{\mathbf k,\mathbf q}$. The <bosonic squeezed-vacuum contractions> are consequently
$$
\boxed{\begin{aligned}
\langle\psi_{\mathbf k}^\dagger\psi_{\mathbf q}\rangle
&=s_k^2\delta_{\mathbf k,\mathbf q},\\
\langle\psi_{\mathbf k}\psi_{\mathbf q}\rangle
&=-c_ks_k\delta_{\mathbf k,-\mathbf q},\\
\langle\psi_{\mathbf k}^\dagger\psi_{\mathbf q}^\dagger\rangle
&=-c_ks_k\delta_{\mathbf k,-\mathbf q}.
\end{aligned}}
$$
In particular the three requested quantities are $-\sinh\theta_k\cosh\theta_k$, $-\sinh\theta_k\cosh\theta_k$ and $\sinh^2\theta_k$, respectively. The minus sign in the pair <vacuum expectation values> comes from the inverse <Bogoliubov transformation> and is essential to lowering the <interaction energy>. Substitution yields
$$
\boxed{\langle H_{\rm int}\rangle
=\frac{uN_0^2}{2V}+\frac{uN_0}{V}\sum_{\mathbf k\ne0}(2s_k^2-c_ks_k)+O(N_0^0).}
$$
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