= Solution
Use the PDF's notation $f_0$ for the source-frame pulse <frequency>, and set $\beta=v/c$, $\gamma=(1-\beta^2)^{-1/2}$. Successive emissions occur at the same train-frame position and are separated by <proper time> $1/f_0$. The <Lorentz transformation> gives a station-frame emission interval $\Delta t_e=\gamma/f_0$ and displacement $v\Delta t_e$. For a receding beacon, the second pulse travels an extra distance $v\Delta t_e$, so
$$
\Delta t_r=\Delta t_e(1+\beta),\qquad \boxed{\frac f{f_0}=\frac1{\gamma(1+\beta)}=\sqrt{\frac{1-\beta}{1+\beta}}.}
$$
This derivation includes both <time dilation> and the change in light-travel time. Expanding for $\beta\ll1$ gives $f/f_0=1-\beta+O(\beta^2)$, recovering the stated nonrelativistic approximation.
Let $\lambda_0$ be the common <wavelength> of the identical lights at rest. The approaching and receding <relativistic Doppler effect> factors give
$$
\lambda_1=\lambda_0\sqrt{\frac{1-\beta}{1+\beta}},\qquad \lambda_2=\lambda_0\sqrt{\frac{1+\beta}{1-\beta}}.
$$
Their ratio and product therefore determine the <speed> and rest <wavelength> separately:
$$
\boxed{\beta=\frac{\lambda_2-\lambda_1}{\lambda_2+\lambda_1}=\frac{636-530}{636+530}=\frac1{11},\qquad v=0.0909c,}
$$
$$
\boxed{\lambda_0=\sqrt{\lambda_1\lambda_2}=580.6\,\mathrm{nm}.}
$$
No assumption about the rest color was needed.
The last observation requires a transverse-geometry convention. The general observer-frame ray formula is
$$
\boxed{\lambda=\gamma\lambda'_3(1-\beta\cos\theta),}
$$
where $\theta$ is the angle between source <velocity> and photon propagation \b[in the platform frame at emission]. Under the usual transverse-ray interpretation, $\theta=\pi/2$, so for $\beta=0.6$ and $\gamma=1.25$,
$$
\boxed{\lambda=\gamma\lambda'_3=500\,\mathrm{nm}\quad\text{for a platform-frame transverse ray}.}
$$
This is the standard transverse redshift from <time dilation>.
There is a genuine distinction between emission and reception at the instant of passing. A side-mounted source emitting perpendicularly in its own frame instead has $\cos\theta=\beta$ by <aberration of light>, giving $\lambda=\lambda'_3/\gamma=320\,\mathrm{nm}$. The same blue result occurs if “at the exact moment” means reception when the source's current position is at closest approach, with a small finite transverse offset. For offset $d$, reception and emission satisfy $t_r=t_e+\sqrt{v^2t_e^2+d^2}/c$. Setting $t_r=0$ gives $t_e=-\gamma d/c$ and $\cos\theta=\beta$; setting $t_e=0$ instead gives the transverse ray received at $t_r=d/c$. These limits remain different as $d\to0$.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2017-doppler-geometry.png]
{title=Emission at closest approach and reception at closest approach give different Doppler factors}
Thus \b[500 nm requires the transverse ray in the platform frame]; \b[320 nm applies to reception at closest approach or transverse emission in the train frame]. The wording does not distinguish these event/frame choices, so merely neglecting the source-observer separation is insufficient to select between them. The reusable derivation is <transverse Doppler shift and observation time>.
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