= Solution
Let $\Psi_3=A^3\psi_3$ and $\Theta_3=A^3\theta_3$ denote the full third-order coefficients. This notation keeps the detuning term, proportional to $A$, and the slow-time term, proportional to $A_T$, visible. Define
$$
B=J(\psi_1,\theta_2)+J(\psi_2,\theta_1).
$$
The equations at order $\epsilon^3$ are
$$
\nabla^2\Psi_3+R_c\Theta_{3x}=-A\theta_{1x},\qquad
\nabla^2\Theta_3-\Psi_{3x}=A_T\theta_1+A^3B.
$$
Equivalently, where $A\ne0$, the question's coefficients satisfy $\nabla^2\psi_3+R_c\theta_{3x}=-A^{-2}\theta_{1x}$ and $\nabla^2\theta_3-\psi_{3x}=(A_T/A^3)\theta_1+B$. Eliminate $\Theta_3$ from the full coefficients:
$$
(\nabla^4+R_c\partial_x^2)\Psi_3=-A\partial_x\nabla^2\theta_1-R_cA_T\theta_{1x}-R_cA^3\partial_xB.
$$
Use the <Fredholm solvability condition for a self-adjoint operator>. With $\langle f,g\rangle=\int_0^1\int_0^2fg\,dx\,dz$, projection onto the critical <eigenfunction> $\psi_1$ makes the left side vanish. Therefore the <Landau amplitude equation> is
$$
\boxed{A_T=\alpha A-\beta A^3,\qquad
\alpha=-\frac{\langle\psi_1,\partial_x\nabla^2\theta_1\rangle}{R_c\langle\psi_1,\theta_{1x}\rangle},\qquad
\beta=\frac{\langle\psi_1,\partial_x\{J(\psi_1,\theta_2)+J(\psi_2,\theta_1)\}\rangle}{\langle\psi_1,\theta_{1x}\rangle}.}
$$
These integrals already answer the requested coefficient calculation. Evaluating them gives $\langle\psi_1,\theta_{1x}\rangle=1/4$, $\langle\psi_1,\partial_x\nabla^2\theta_1\rangle=-\pi^2/2$, and $\langle\psi_1,\partial_xB\rangle=\pi^2/32$. Hence, for the prescribed normalization,
$$
\boxed{A_T=\frac12A-\frac{\pi^2}{8}A^3.}
$$
The cubic term opposes growth, so the nonzero steady amplitudes $A=\pm2/\pi$ are <stable equilibria> within the chosen roll phase. The conductive state $A=0$ is unstable for this positive detuning. This is <supercritical saturation of Darcy convection rolls>.
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