= Solution
The lower-layer <dispersion relation> gives
$$
\boxed{m_0=k\sqrt{N_0^2/\omega^2-1}.}
$$
The incident phase is $e^{-im_0z}$; its vertical <group velocity> is positive, even though its vertical phase velocity is negative. Write the lower field as $W=w_0e^{-im_0z}+r e^{im_0z}$ and the upper bounded field as $W=T e^{-kz}$. Interface continuity gives $T=w_0+r$ and
$$
-kT+im_0(w_0-r)=-kS T,\qquad S=\Delta b k/\omega^2.
$$
Thus the <internal-wave response of a sharp buoyancy interface> is
$$
\frac{T}{w_0}=\frac{2im_0}{k(1-S)+im_0},\qquad
\frac{|T|^2}{|w_0|^2}=\frac{4m_0^2}{k^2(1-S)^2+m_0^2}.
$$
At fixed incident $k,\omega,N_0$ this is largest at \b[$S=1$], where $|T|=2|w_0|$. Consequently $\omega^2=\Delta b k=2\omega_{\rm free}^2$, or \b[$\omega=\sqrt2\,\omega_{\rm free}$]. The maximum does not coincide with the free frequency for two homogeneous layers: the stratified lower radiating layer has a different dynamic pressure response. The upper field is <evanescent>, so in this ideal lossless problem it carries no mean outgoing vertical <energy flux>; the reflected wave has $|r|=|w_0|$.
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