= Solution
Use <conformal time>, $dt=a\,d\tau$, and let primes mean $d/d\tau$. Then $a'=a\dot a$ and $a''=a\dot a^2+a^2\ddot a$. For $k=1$, $\Lambda=0$, the <Friedmann equation> and acceleration equation give
$$
\dot a^2=\frac{\rho_0}{3}a^{-(1+3w)}-1,\qquad
\ddot a=-\frac{\rho_0}{6}(1+3w)a^{-(2+3w)}.
$$
Substitution yields
$$
\boxed{a''+a=\frac{\rho_0}{6}(1-3w)a^{-3w}}.
$$
Use the supplied solution without deriving it. Its first maximum has sine equal to one, so its amplitude must equal the physical turnaround scale in part (a):
$$
\boxed{A=\left(\frac{\rho_0}{3}\right)^{1/(1+3w)}}.
$$
Choose the expanding branch with $B=0$ and the first zero at $\tau=0$.
For <pressureless matter>, $w=0$, the <closed matter-dominated Friedmann solution> is
$$
\boxed{a(\tau)=A\sin^2(\tau/2)=\frac A2(1-\cos\tau),\quad
t(\tau)=\frac A2(\tau-\sin\tau),\quad A=\rho_0/3}.
$$
The <Big Crunch> is at $\tau_c=2\pi$, or <proper time> $t_c=\pi A=\pi\rho_0/3$, measured from the <Big Bang>.
For radiation, $w=1/3$, the <closed radiation-dominated Friedmann solution> is
$$
\boxed{a(\tau)=A\sin\tau,\quad t(\tau)=A(1-\cos\tau),\quad A=\sqrt{\rho_0/3}}.
$$
The <Big Crunch> is at $\tau_c=\pi$, or $t_c=2A$.
A <radial null geodesic in FLRW spacetime> obeys $d\chi/d\tau=\pm1$. The spatial slice is a unit three-sphere times $a$, so a great circle has comoving circumference $2\pi$. The dust lifetime supplies conformal distance $2\pi$: \b[one circumference, with the return occurring only in the crunch limit]. Strictly before the singular endpoint no full return is completed. The radiation lifetime supplies conformal distance $\pi$: \b[the <photon> reaches the antipode, half a great circle, in the crunch limit]. These are limiting null rays from near the initial singularity, not <photons> at a regular event on the singular surface.
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