= Solution
Substitution of a plane wave in the <Schrödinger equation> gives $\hbar\omega=\hbar^2k^2/(2m)+V_0$. For a stationary step-scattering state, remove the common phase $e^{-iEt/\hbar}$ and write $\psi_L=e^{ikx}+r e^{-ikx}$, $k=\sqrt{2mE}/\hbar$. If $E>V_0$, the right-side outgoing wave is $\psi_R=t e^{iqx}$, $q=\sqrt{2m(E-V_0)}/\hbar$. A finite step with no delta-function potential in the Schrödinger operator requires continuous $\psi$ and $\psi'$: a discontinuous <wavefunction> would generate a delta derivative, and integrating the equation over a vanishing interval gives derivative continuity. Matching yields
$$
r=\frac{k-q}{k+q},\qquad t=\frac{2k}{k+q},\qquad
\boxed{\mathcal R=\left(\frac{\sqrt E-\sqrt{E-V_0}}{\sqrt E+\sqrt{E-V_0}}\right)^2\quad(E>V_0).}
$$
For $0<E<V_0$, take $q=i\kappa$ and the bounded right-side state $te^{-\kappa x}$. It carries no transmitted flux and $\mathcal R=1$. The threshold limit also has reflection <probability> one. Thus at $V_0=1\,\mathrm{eV}$,
$$
\boxed{\mathcal R(0.9\,\mathrm{eV})=1,\qquad\mathcal R(1.1\,\mathrm{eV})\simeq0.288.}
$$
For the two-step profile, set $k_0=\sqrt{2mE}/\hbar$, $k_1=\sqrt{2m(E-V_1)}/\hbar$ and $k_2=\sqrt{2m(E-V_0)}/\hbar$. With all regions propagating, the net reflection amplitude is
$$
r_{\rm tot}=\frac{r_{01}+r_{12}e^{2ik_1a}}{1+r_{01}r_{12}e^{2ik_1a}},\qquad
r_{ij}=\frac{k_i-k_j}{k_i+k_j}.
$$
The given quarter-wave condition makes $e^{2ik_1a}=-1$. Zero reflection then requires $r_{01}=r_{12}$, equivalent to $k_1^2=k_0k_2$. Consequently
$$
\boxed{E-V_1=\sqrt{E(E-V_0)},\qquad\text{or}\qquad E=\frac{V_1^2}{2V_1-V_0}.}
$$
For an increasing intermediate step this needs $E>V_0$ and $V_0/2<V_1<V_0$, together with the stipulated thickness condition. In this <quarter-wave matching layer for a quantum potential step>, two reflected amplitudes cancel while the transmitted flux equals the incident flux.
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