= Solution
Write $\omega$ for the nonzero forcing <frequency>, to distinguish it from the vertical <velocity> $w$. Put $k_z=\pi/h$ and $Z(z)=\cos(\pi(2z+h)/(2h))=-\sin(k_zz)$. Then $Z(0)=Z(h)=0$ and $Z''=-k_z^2Z$, so it meets the rigid-wall conditions. Substituting $w=\operatorname{Re}(f(x)e^{-i\omega t}Z(z))$ into the scalar <wave> equation gives
$$
-\omega^2(f''-k_z^2f)-4\Omega^2k_z^2f=0,
$$
or
$$
\boxed{f''+k_z^2\left(\frac{4\Omega^2}{\omega^2}-1\right)f=0,\qquad f(0)=a.}
$$
The <radiation sign for an inertial wave> must exclude any disturbance arriving from $x=+\infty$. This means an <radiation condition> for propagating <waves>, or bounded decay for an evanescent <wave>.
For $0<|\omega|<2|\Omega|$, put $q=k_z\sqrt{4\Omega^2/\omega^2-1}$. The solutions are oscillatory. The <inertial-wave dispersion relation> is $\omega^2=4\Omega^2k_z^2/(k_x^2+k_z^2)$; its horizontal <group velocity> is $c_{g,x}=-\omega k_x/(k_x^2+k_z^2)$. Positive energy transport into the channel therefore requires $k_x=-\operatorname{sgn}(\omega)q$. The solution is
$$
\boxed{w(x,z,t)=\operatorname{Re}\left[a\,e^{-i\operatorname{sgn}(\omega)qx-i\omega t}\cos\left(\frac\pi{2h}(2z+h)\right)\right].}
$$
For a positive <frequency>, the phase travels towards negative $x$ while the <group velocity> travels towards positive $x$. Choosing the outgoing sign from the <phase velocity> alone would give the wrong result.
For $|\omega|>2|\Omega|$, set $\kappa=k_z\sqrt{1-4\Omega^2/\omega^2}$. Excluding the exponentially growing solution gives
$$
\boxed{w(x,z,t)=\operatorname{Re}\left[a\,e^{-\kappa x-i\omega t}\cos\left(\frac\pi{2h}(2z+h)\right)\right].}
$$
This disturbance decays away from the wavemaker and does not propagate horizontally. At the cutoff the ODE degenerates to $f''=0$; zero <frequency> is not a nonzero forced harmonic mode with this vertical shape.
There is a second printed inconsistency: the stated <velocity> $u\mathbf i+w\mathbf k$, if taken literally with no $y$ dependence and no transverse <velocity>, cannot support a nontrivial rotating <inertial wave>. Its transverse <momentum> equation would give $2\Omega u=0$, then continuity and the rigid walls would <force> $w=0$. The scalar solutions above are the intended inertial-<wave> solutions when two-dimensional means dependence on $x,z$ while \b[retaining the transverse component $v$]. For a mode, $-i\omega v+2\Omega u=0$ explicitly supplies the generally nonzero transverse <velocity>. Thus the scalar equation can be solved as requested, but the full <vector> field needs this correction to the printed two-component description.
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