= Solution
The <Time-independent Schrödinger equation> is the <energy eigenvalue> equation $\widehat H\psi=E\psi$ for a particle of <mass> $m$. The <Hamiltonian operator> is the sum of the kinetic operator $-(\hbar^2/2m)d^2/dx^2$ and the prescribed <potential energy> $V(x)$. Here $\hbar$ is the <reduced Planck constant>. A corresponding <stationary state> is $\Psi(x,t)=\psi(x)e^{-iEt/\hbar}$. The <Born rule> identifies $|\Psi|^2=|\psi|^2$ as <probability density> for a state with <wavefunction normalization> $\int|\Psi|^2dx=1$; integrating it over an interval gives a position <probability>. A <plane wave> scattering state is not a <normalizable wavefunction> on the whole line, but its <probability currents> give the reflection and transmission fractions of an incident <wave packet>.
There is a notation discrepancy in the sources: the original PDF prints an unbarred $h^2$ in the kinetic term, whereas the local TeX uses $\hbar^2$. The standard physical equation uses the <reduced Planck constant> $\hbar=h/(2\pi)$, where $h$ is the <Planck constant>; that convention is used below. Reading the printed $h$ literally as the <Planck constant> would scale all lengths in this model by $2\pi$ relative to the standard answers, but would leave the amplitude ratios unchanged.
For $E>V_0$, both sides allow oscillatory <wavefunctions>. A classical particle always crosses the <potential step> and slows because its <kinetic energy> falls from $E$ to $E-V_0$. In <quantum mechanics> there is generally both reflection and transmission even though the particle is above the step. Define the positive <wavenumbers>
$$
k=\frac{\sqrt{2mE}}\hbar,\qquad q=\frac{\sqrt{2m(E-V_0)}}\hbar.
$$
With no wave incident from the right, write
$$
\psi(x)=\begin{cases}Ae^{ikx}+Be^{-ikx},&x\le0,\\Ce^{iqx},&x>0.\end{cases}
$$
A finite <potential step> with constant <mass> makes both the <wavefunction> and its first <derivative> continuous. The two <boundary conditions> at $x=0$ are
$$
A+B=C,\qquad k(A-B)=qC.
$$
Solving gives the reflected and transmitted <wavefunctions>
$$
\boxed{\psi_{\rm ref}(x)=A\frac{k-q}{k+q}e^{-ikx},\qquad\psi_{\rm tr}(x)=A\frac{2k}{k+q}e^{iqx}.}
$$
For the specified <Electron> energies, $q/k=\sqrt{10/50}=1/\sqrt5$, so the relative <probability amplitudes> are
$$
\boxed{\frac BA=\frac{\sqrt5-1}{\sqrt5+1}=0.381966,\qquad\frac CA=\frac{2\sqrt5}{\sqrt5+1}=1.381966.}
$$
The transmitted <probability amplitude> exceeding one is compatible with <conservation of quantum probability>, since the outgoing <speed> is smaller. In fact the <probability current> $j=(\hbar/m)\operatorname{Im}(\psi^*\psi')$ gives
$$
\boxed{\mathcal R=\left|\frac BA\right|^2=0.145898,\qquad\mathcal T=\frac qk\left|\frac CA\right|^2=0.854102,\qquad\mathcal R+\mathcal T=1.}
$$
These are the <quantum reflection probability> and <transmission probability>, which must be distinguished from the requested relative <amplitudes>.
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