= Solution
Assume the atmosphere is transparent to incoming shortwave radiation, as required for the stated infrared-only radiative-equilibrium model. Global <energy> balance gives
$$
\boxed{L_\infty=\frac{(1-\alpha)L_{\circledast}}4.}
$$
Put $D=\tau_\infty$ and $F=I_+-I_-$. <Radiative equilibrium> requires $dF/d\tau=0$, hence the two equations imply $I_++I_-=2\sigma T^4$. Substitution gives $dI_+/d\tau=dI_-/d\tau=-F/2$. At the top there is no incident infrared flux, $I_-(D)=0$, and $I_+(D)=F=L_\infty$. Therefore
$$
\boxed{I_+(\tau)=L_\infty\left[1+\frac{D-\tau}{2}\right],\qquad
I_-(\tau)=\frac{L_\infty}{2}(D-\tau).}
$$
The black ground supplies $I_+(0)=\sigma T_g^4$. Thus
$$
\boxed{L_\infty=\frac{2\sigma T_g^4}{2+\tau_\infty}.}
$$
The ground also satisfies $\sigma T_g^4=L_\infty+I_-(0)$, showing explicitly that its outgoing flux balances absorbed sunlight plus downward atmospheric radiation. In pure <radiative equilibrium> the air just above the ground need not have the ground's <temperature>: $\sigma T(0)^4=L_\infty(1+D)/2$, whereas $\sigma T_g^4=L_\infty(2+D)/2$. That discontinuity is relevant to surface-driven <convection> in part (d).
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