= Solution
Here $V(x)$ is <potential energy>, $E$ is the stationary state's total <energy>, and $\psi(x)$ is its spatial <probability amplitude>. The <wave function> has time factor $e^{-iEt/\hbar}$; $|\psi|^2$ is a position <probability density> for a normalized state. Stationary plane-wave scattering solutions are idealizations described by probability-current ratios, or components of a normalizable <wave packet>.
Above an upward step, a classical particle is fully transmitted with reduced <speed>. Quantum mechanically the abrupt <wavelength> change generally produces both reflected and transmitted <amplitudes> even though both regions are classically allowed. With unit incident <amplitude> and no <wave> arriving from the right,
$$
\psi(x)=\begin{cases}e^{ik_1x}+r e^{-ik_1x},&x<0,\\t e^{ik_2x},&x>0,\end{cases}\qquad k_1=\frac{\sqrt{2mE}}{\hbar},\quad k_2=\frac{\sqrt{2m(E-V_0)}}{\hbar}.
$$
The reflected component has signed wavevector $-k_1$ and the same <wavenumber> magnitude as the incident component. A finite potential jump with constant <mass> and no delta interaction requires continuity of $\psi$ and $\psi'$:
$$
1+r=t,\qquad k_1(1-r)=k_2t,\qquad\boxed{r=\frac{k_1-k_2}{k_1+k_2},\quad t=\frac{2k_1}{k_1+k_2}.}
$$
The <probability current> $j=(\hbar/m)\operatorname{Im}(\psi^*\psi')$ gives probability reflection and transmission $\mathcal R=|r|^2$ and $\mathcal T=(k_2/k_1)|t|^2$, with $\mathcal R+\mathcal T=1$. The <amplitude> $t$ may exceed one without violating conservation of total probability, because the transmitted group <speed> is smaller.
For $0<E<V_0$, let $\kappa=\sqrt{2m(V_0-E)}/\hbar$. The bounded right-hand solution is $t e^{-\kappa x}$, an <evanescent wave> tail with <amplitude> penetration scale $1/\kappa$. It carries no outgoing <probability current> in an infinitely long forbidden half-space, so reflection is total. This differs from transmission through a finite <potential barrier>, where a second boundary can support <quantum tunnelling>.
For a downward step, replace $k_2$ by $\sqrt{2m(E+V_0)}/\hbar>k_1$. The same <boundary conditions> give a nonzero negative $r$, corresponding to a reflected <amplitude> with <phase> reversal. If reflection were forbidden, $r=0$ would require both $t=1$ and $k_1=k_2$, impossible for a nonzero step. Reflection therefore does not require a classical <potential barrier>; the discontinuity of <wavenumber> mismatch is enough.
For the specified <electron> energies, $k_2/k_1=\sqrt{40/100}=\sqrt{0.4}$, so the requested <amplitude> coefficients are
$$
\boxed{r=0.22515,\qquad t=1.22515.}
$$
The corresponding probability coefficients are approximately $\mathcal R=0.05069$ and $\mathcal T=0.94931$. <electron mass> cancels from these ratios; there is no need to invent an additional material <effective mass>.
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