= Solution
Assume a positive-energy incoming particle, $E>mc^2$, and a nonzero upward step $V_0>0$. Set $M=mc^2$ and $k=\sqrt{E^2-M^2}/(\hbar c)>0$. With the common time factor $e^{-iEt/\hbar}$ removed, take $\phi_<(x)=e^{ikx}+r e^{-ikx}$ and $\phi_>(x)=t e^{iqx}$. The right-hand wavenumber satisfies $q^2=[(E-V_0)^2-M^2]/(\hbar c)^2$; its physical branch is specified in part (c). Integrating the stationary equation across the finite step requires continuity of $\phi$ and $\phi'$, giving $1+r=t$ and $k(1-r)=qt$. Thus
$$
\boxed{r=\frac{k-q}{k+q},\qquad t=\frac{2k}{k+q}}.
$$
These are wave amplitudes, not <flux> probabilities. The <Noether current of scalar electrodynamics>, in a convenient common normalization for a stationary mode, is $\rho_Q=(E-V)|\phi|^2/(mc^2)$ and $j_Q=(\hbar/m)\operatorname{Im}(\phi^*\partial_x\phi)$. For a real $q$, its charge-current ratios are $\mathcal R=|r|^2$ and $\mathcal T=(q/k)|t|^2$, with $\mathcal R+\mathcal T=1$. The same statement uses $\operatorname{Re}q$ for a decaying solution. The <electrostatic potential> couples through the time derivative; it is not a Lorentz-scalar <mass> shift. A zero step is recovered separately with $q=k$, $r=0$ and $t=1$.
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