= Solution
Separate continuity gives
$$
\rho_r=\rho_{r0}a^{-4}
=\beta\frac{3c^2H_0^2}{8\pi G}a^{-4},
\qquad
\rho_\Lambda=\rho_{\Lambda0}
=\frac{3c^2H_0^2}{8\pi G},
$$
using the scalings for <radiation in cosmology> and <cosmological constant energy>. At $t=t_0$, where $a=1$, the Friedmann equation becomes
$$
H_0^2+Kc^2=(\beta+1)H_0^2,
$$
so
$$
\boxed{Kc^2=\beta H_0^2}.
$$
At general $a$,
$$
H^2+\frac{\beta H_0^2}{a^2}
=H_0^2\left(\frac\beta{a^4}+1\right),
$$
which rearranges to
$$
\boxed{
H^2=\frac{H_0^2}{a^4}(\beta-\beta a^2+a^4)
}.
$$
Part (a), applied to $\rho=\rho_r+\rho_\Lambda$ and $P=\rho_r/3-\rho_\Lambda$, gives
$$
\dot H+H^2
=-H_0^2\left(\frac\beta{a^4}-1\right).
$$
Subtracting the displayed formula for $H^2$ yields
$$
\boxed{
\dot H=-\beta\frac{H_0^2}{a^4}(2-a^2)
}.
$$
These relations define the <closed radiation--cosmological-constant turning polynomial>.
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