= Solution
\b[No: ordering the attractive potentials does not impose the proposed ordering of reflection probabilities.] Take $k=\pi/(4a)$ and let $M$ denote the particle mass. Choose $V_2=0$ everywhere and
$$
V_1(x)=\begin{cases}-4\hbar^2k^2/M,&|x|<a,\\0,&|x|\ge a.\end{cases}
$$
These satisfy every stated ordering and support condition. The incoming energy is $\hbar^2k^2/(2M)$; the interior <wavenumber> in the <finite square well> is $q=\sqrt{k^2-2MV_1/\hbar^2}=3k$.
Here is an explicit matching calculation. For a general well of width $L=2a$, take exterior waves $e^{ik(x+a)}+re^{-ik(x+a)}$ on the left and $te^{ik(x-a)}$ on the right. Shifting the incident phase in this way does not change the <reflection probability>. Continuity of the <wavefunction> and its first <derivative> gives the propagation relation
$$
\begin{pmatrix}t\\ikt\end{pmatrix}
=\begin{pmatrix}\cos(qL)&\sin(qL)/q\\-q\sin(qL)&\cos(qL)\end{pmatrix}
\begin{pmatrix}1+r\\ik(1-r)\end{pmatrix}.
$$
Inverting it and adding the two equations after dividing the <derivative> equation by $ik$ yields
$$
t=\frac2{2\cos(qL)-i(k/q+q/k)\sin(qL)},\qquad
r=\frac{i(q/k-k/q)\sin(qL)}{2\cos(qL)-i(k/q+q/k)\sin(qL)}.
$$
The incident and reflected exterior waves have the same speed, so their <flux> ratio is
$$
p_1=|r|^2=\frac{(q^2-k^2)^2\sin^2(qL)}{4k^2q^2+(q^2-k^2)^2\sin^2(qL)}.
$$
For the chosen parameters $qL=3\pi/2$ and $(q^2-k^2)^2/(4k^2q^2)=16/9$. Thus
$$
\boxed{p_1=\frac{16}{25},\qquad p_2=0.}
$$
The zero potential has no reflected component. This proves the failure with explicit probabilities. More generally the same expression vanishes at $qL\in\pi\mathbb Z$, explaining <resonant transmission through a square well> and why increasing attractive well depth does not monotonically control reflection.
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