= Solution
Specify the <Fourier transform> before performing the <Gaussian integral>:
$$
m_j(\mathbf q)=\int d^3x\,e^{-i\mathbf q\cdot\mathbf x}m_j(\mathbf x),\qquad
m_j(\mathbf x)=\int\frac{d^3q}{(2\pi)^3}e^{i\mathbf q\cdot\mathbf x}m_j(\mathbf q).
$$
At $u=0$ the functional is diagonal in both component and momentum, with inverse kernels $q^2+t+r$, $q^2+t-r$, and $q^2+t$. The condition $0<r<t$ makes all three positive. The <covariance> of a <Gaussian distribution> with statistical weight $e^{-F/(k_BT)}$ is $k_BT$ times the inverse kernel, hence
$$
\boxed{\begin{aligned}
\langle m_1(\mathbf q)m_1(\mathbf p)\rangle&=(2\pi)^3\delta^{(3)}(\mathbf q+\mathbf p)\frac{k_BT}{q^2+t+r},\\
\langle m_2(\mathbf q)m_2(\mathbf p)\rangle&=(2\pi)^3\delta^{(3)}(\mathbf q+\mathbf p)\frac{k_BT}{q^2+t-r}.
\end{aligned}}
$$
If $F$ is already the dimensionless <free energy> divided by $k_BT$, the numerators are one instead. Translational invariance forces opposite momenta to pair. There is no complex conjugate on the second field, so the delta function is at $\mathbf p=-\mathbf q$, not at $\mathbf p=\mathbf q$. A correlation with $m_j(\mathbf p)^*$ would have the latter convention. Rotational symmetry in real space leaves dependence on $q^2$ alone once the momentum-conservation delta function is factored out. The <correlation lengths> are $\xi_1=(t+r)^{-1/2}$ and $\xi_2=(t-r)^{-1/2}$; the second component becomes soft as $r\uparrow t$. The skipped letter in the source is preserved: this solution is part (e), with no inferred part (d).
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