= Solution
Use the <Wirtinger derivative> $q_z=(q_x-iq_y)/2$ and write $z=x+iy$. Since $q$ solves the <Laplace equation>, $q_z$ is a <holomorphic function>. The transforms in the question select solutions with sufficient decay at infinity. In this class the solution, when it exists, is unique; hence reflection of the symmetric data about $y=\ell/2$ gives
$$
q(x,\ell-y)=q(x,y),\qquad q(x,\ell)=q(x,0).
$$
For $\gamma>0$, this uniqueness follows directly from <Green's first identity>: the homogeneous problem has integral $\int|\nabla q|^2+\gamma\int_{y=0,\ell}q^2=0$. For arbitrary real $\gamma$, the decaying-class qualifications are explained in part (c). Without a condition at infinity, symmetry of the data alone would not force symmetry of every solution.
Put $c=q(0,0)$, $f(x)=q(x,0)$, and introduce a known transform
$$
\psi(s)=\frac12\int_0^\infty e^{sx}f(x)dx,\qquad H(k)=\frac12\int_0^\ell e^{ky}g(y)dy.
$$
The bottom <Robin boundary condition> gives $q_y=\gamma f$, while the top one gives $q_y=-\gamma f$. The bottom side is traversed from infinity to zero. <Integration by parts> therefore gives
$$
G_1=-\frac12\int_0^\infty e^{-ikx}(f'-i\gamma f)dx=\boxed{\frac c2-i(k-\gamma)\psi(-ik)}.
$$
The top side is traversed from $i\ell$ towards infinity, and its exponential supplies $e^{k\ell}$, giving
$$
\boxed{G_3=e^{k\ell}\left[-\frac c2+i(k+\gamma)\psi(-ik)\right].}
$$
On the vertical side, $dz=i\,dy$ and the prescribed <Neumann boundary condition> is $q_x=g$. Thus
$$
G_2=\frac12\int_0^\ell e^{ky}(q_y+ig)dy=\boxed{\phi(k)+iH(k)}.
$$
Both the side orientations and the factor $1/2$ from the <Wirtinger derivative> are essential to these signs.
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