= Solution
Let $S$ denote specific entropy, reserving $s$ from the first question for envelope thickness. The <stellar adiabatic exponents> are defined at fixed composition by
$$
\boxed{\Gamma_1=\left(\frac{\partial\log P}{\partial\log\rho}\right)_S,
\qquad \frac{\Gamma_2-1}{\Gamma_2}
=\left(\frac{\partial\log T}{\partial\log P}\right)_S.}
$$
It is useful to introduce $\Gamma_3-1=(\partial\log T/\partial\log\rho)_S$, so $(\Gamma_2-1)/\Gamma_2=(\Gamma_3-1)/\Gamma_1$.
Take the stellar gas to be a fully ionized, nonrelativistic monatomic <ideal gas>, with fixed specific gas constant $\mathcal R$. Its mixture with thermal radiation has
$$
P=\mathcal R\rho T+\frac13a_{\rm r}T^4,\qquad
u=\frac32\mathcal RT+\frac{a_{\rm r}T^4}{\rho}.
$$
Here $u$ is specific internal energy, and the <stellar gas-pressure fraction> is $\beta=\mathcal R\rho T/P$. The assumption of monatomic gas fixes its heat capacity; the perfect-gas <pressure> law alone would not determine it.
At fixed $T$ and fixed $\rho$, respectively, the <pressure> derivatives are
$$
\chi_\rho=\left(\frac{\partial\log P}{\partial\log\rho}\right)_T=\beta,
\qquad
\chi_T=\left(\frac{\partial\log P}{\partial\log T}\right)_\rho=4-3\beta.
$$
For an <isentropic process>, the <first law of thermodynamics> gives $du=P\,d\rho/\rho^2$. Differentiate the internal energy rather than artificially holding $\beta$ fixed during the perturbation:
$$
\left(\frac32\mathcal R+\frac{4a_{\rm r}T^3}{\rho}\right)dT
=\frac{P+a_{\rm r}T^4}{\rho^2}d\rho.
$$
The parenthesis is $(P/\rho T)[(3/2)\beta+12(1-\beta)]$, giving
$$
\Gamma_3-1=\frac{4-3\beta}{12-(21/2)\beta}
=\frac{8-6\beta}{24-21\beta}.
$$
Since $d\log P=\chi_\rho d\log\rho+\chi_Td\log T$,
$$
\boxed{\Gamma_1=\beta+(4-3\beta)(\Gamma_3-1)
=\frac{32-24\beta-3\beta^2}{24-21\beta}.}
$$
Using the relation between the exponents gives the <adiabatic exponents of a monatomic gas-radiation mixture>
$$
\boxed{\Gamma_2=\frac{32-24\beta-3\beta^2}{24-18\beta-3\beta^2},\qquad
\nabla_{\rm ad}=\frac{\Gamma_2-1}{\Gamma_2}
=\frac{8-6\beta}{32-24\beta-3\beta^2}.}
$$
The requested values, including the <adiabatic temperature gradient>, are
$$
\boxed{\begin{array}{c|ccc}
\beta&\Gamma_1&\Gamma_2&\nabla_{\rm ad}\\\hline
0&4/3&4/3&1/4\\
1/2&77/54&77/57&20/77\\
1&5/3&5/3&2/5
\end{array}.}
$$
The pure-radiation and pure-gas rows are understood as limits of the mixture.
For uniform composition, the <Schwarzschild criterion> for <stellar convective instability> is
$$
\boxed{\nabla\equiv\frac{d\log T}{d\log P}>\nabla_{\rm ad}
=\frac{\Gamma_2-1}{\Gamma_2}.}
$$
To test whether a radiative configuration becomes unstable, use its required <stellar radiative temperature gradient> $\nabla_{\rm rad}$ for $\nabla$. Since $dP/dr<0$, the equivalent radial condition is
$$
-\frac{dT}{dr}>\frac{\Gamma_2-1}{\Gamma_2}\frac TP\left(-\frac{dP}{dr}\right).
$$
An outward-displaced parcel then cools less than its new surroundings, remains less dense at the same <pressure>, and is further accelerated outward.
The requested <Eddington-model convective-core mass fraction> follows from the usual global constant-$\beta$ closure. At the outer radiative surface, put $m=M$, $L_r=L$, and $P_{\rm rad}=(1-\beta)P$. Dividing the radiation-pressure gradient by the <stellar hydrostatic equation> gives
$$
\frac{dP_{\rm rad}}{dP}=\frac{\kappa L_r}{4\pi cGm},\qquad
L=\frac{4\pi cGM}{\kappa}(1-\beta)=(1-\beta)L_{\rm Edd}.
$$
There is no energy generation outside the <convective core>, so use $L_r=L$ there. The local radiative-gradient formula is
$$
\nabla_{\rm rad}(m)=\frac{3\kappa L P}{16\pi a_{\rm r}cGmT^4}
=\frac{\kappa L}{16\pi cGm(1-\beta)}\simeq\frac{M}{4m}.
$$
The same representative $\beta$ has been used in the global <luminosity> closure and at the boundary. Setting this gradient equal to $\nabla_{\rm ad}$ at $m=M_c$ produces
$$
\boxed{\frac{M_c}{M}=\frac1{4\nabla_{\rm ad}}
=\frac{\Gamma_2}{4(\Gamma_2-1)}
=\frac{32-24\beta-3\beta^2}{4(8-6\beta)}.}
$$
This is $1$, $77/80$, and $5/8$ for $\beta=0,1/2,1$, respectively, with $\beta=0$ a marginal radiation-dominated limit.
There is a consistency qualification to this last model estimate. \b[It cannot be an exact stellar solution if all the printed assumptions are enforced pointwise.] An exactly uniform nonzero radiation fraction would give $dP_{\rm rad}/dP=1-\beta$ at every radiative point. The exact equations would then require
$$
\frac{L_r}{m}=\frac{4\pi cG}{\kappa}(1-\beta).
$$
If $L_r$ is constant throughout an envelope with positive <mass density>, $m$ increases with radius, so this equality cannot hold there. Equivalently uniform $\beta$ fixes the actual gradient to $1/4$, whereas $\nabla_{\rm ad}>1/4$ for $\beta>0$. The boxed core fraction is the intended Eddington closure and boundary estimate, which relaxes exact constancy of $\beta$ in the detailed envelope. The <constant-beta radiative-envelope obstruction> identifies the missing approximation; it is not legitimate to silently assert exact compatibility.
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