= Solution
Let $G_1=\phi(\mathbf r|\mathbf r_1)$ and $G_2=\phi(\mathbf r|\mathbf r_2)$. <Green's second identity> gives
$$
\int_\Omega[G_1(\Delta+k_0^2)G_2-G_2(\Delta+k_0^2)G_1],dV
=\int_S(G_1\partial_nG_2-G_2\partial_nG_1),dS.
$$
The common <Robin boundary condition> makes the surface integrand vanish. Where $\beta\neq0$, substitute $\partial_nG_i=-\alpha G_i/\beta$; a Dirichlet portion also contributes zero. The delta sources therefore give the <Helmholtz Green-function reciprocity>
$$
\boxed{\phi(\mathbf r_1|\mathbf r_2)=\phi(\mathbf r_2|\mathbf r_1).}
$$
This is a bilinear identity, with no complex conjugation. Assume the boundary-value problem has a unique Green kernel; in an exterior problem use the same outgoing radiation condition for both kernels.
For the weighted operator, assume smooth positive $f$ and set $\mathbf b=\nabla\log f$. Then
$$
L=\frac1f\nabla\cdot(f\nabla)+k_0^2=\Delta+\mathbf b\cdot\nabla+k_0^2.
$$
Its unweighted transpose, obtained by <integration by parts>, is
$$
L^{\mathrm T}u=\Delta u-\nabla\cdot(\mathbf bu)+k_0^2u
=\nabla\cdot[f\nabla(u/f)]+k_0^2u.
$$
The associated boundary identity is
$$
\int_\Omega(uLv-vL^{\mathrm T}u),dV
=\int_S[u\partial_nv-v\partial_nu+(\partial_n\log f)uv],dS.
$$
To cancel it when $v$ satisfies the given Robin condition, the required adjoint kernel satisfies
$$
\boxed{\nabla\cdot\left[f\nabla\left(\frac{\widetilde\psi}{f}\right)\right]+k_0^2\widetilde\psi
=\delta(\mathbf r-\mathbf r_2),}
$$
$$
\boxed{\alpha\widetilde\psi+\beta\left(\partial_n\widetilde\psi-\widetilde\psi\,\partial_n\log f\right)=0.}
$$
Using these two delta sources in the boundary identity proves $\psi(\mathbf r_2|\mathbf r_1)=\widetilde\psi(\mathbf r_1|\mathbf r_2)$, the <adjoint reciprocity for a weighted Helmholtz operator>. As a check, $L^{\mathrm T}(fv)=fLv$, so
$$
\widetilde\psi(\mathbf r|\mathbf r_2)=\frac{f(\mathbf r)}{f(\mathbf r_2)}\psi(\mathbf r|\mathbf r_2).
$$
This is equivalent to <weighted acoustic Green-function reciprocity>, $f(\mathbf r_2)\psi(\mathbf r_2|\mathbf r_1)=f(\mathbf r_1)\psi(\mathbf r_1|\mathbf r_2)$. Unit delta normalization is essential here. A Hermitian adjoint would instead conjugate coefficients and change the radiation convention; the requested equality uses the transpose.
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