= Solution
At zero applied field, <time-reversal symmetry> interchanges $M$ and $-M$ without changing the <free energy>, so the analytic expansion contains only even <powers>. The term $-\mu_0HM$ breaks this symmetry at fixed nonzero H. Stationarity and local stability require
$$
\boxed{\mu_0H=a(T)M+bM^3,\qquad a(T)+3bM^2>0.}
$$
For $T>T_c$ at small field, the cubic term is higher order and $M\simeq\mu_0H/a(T)$. Thus the susceptibility as defined in the question is
$$
\boxed{\chi=\left(\frac{\partial M}{\partial H}\right)_T\xrightarrow[H\to0]{}\frac{\mu_0}{a(T)}.}
$$
The three requested free-energy sketches and their minima are given separately above.
For the plotted isotherms, divide the <equation of state> by nonzero M and rearrange:
$$
\boxed{M^2=\frac{\mu_0}{b}\frac HM-\frac{a(T)}b.}
$$
This is an <Arrott plot>: its slope determines b, and the isotherm through the origin identifies the mean-field <Curie temperature>.<Criterion for Ferromagnetism from Observations of Magnetic Isotherms by Anthony Arrott>[1957, abstract: molecular-field criterion from cubic magnetisation versus internal field at the Curie temperature]{ref} Under the paper's stated uniform-field Landau model, use H as supplied on its axis.
The ordinate unit is $(10^4\,\mathrm{A\,m^{-1}})^2=10^8\,\mathrm{A^2\,m^{-2}}$. Reading the original graph gives slopes about $0.076$–$0.080$ ordinate units per unit H/M, so the physical slope is approximately $7.8\times10^6\,\mathrm{A^2\,m^{-2}}$. Consequently
$$
\boxed{b\simeq1.6\times10^{-13}\,\mathrm{J\,m\,A^{-4}}.}
$$
Approximate horizontal intercepts and the corresponding $a=\mu_0(H/M)_{M^2=0}$ are:
* $4.2\,\mathrm K$: intercept $-62$, giving $a\simeq-7.8\times10^{-5}\,\mathrm{H\,m^{-1}}$.
* $13\,\mathrm K$: intercept $-40$, giving $a\simeq-5.1\times10^{-5}\,\mathrm{H\,m^{-1}}$.
* $19\,\mathrm K$: intercept $-15$, giving $a\simeq-1.9\times10^{-5}\,\mathrm{H\,m^{-1}}$.
* $25\,\mathrm K$: intercept $+17$, giving $a\simeq+2.2\times10^{-5}\,\mathrm{H\,m^{-1}}$.
The sign change brackets the transition between 19 and 25 K. A linear fit of $a(T)=\alpha(T-T_c)$ across the four read-offs gives approximately $\alpha=4.8\times10^{-6}\,\mathrm{H\,m^{-1}\,K^{-1}}$ and
$$
\boxed{T_c\simeq22\,\mathrm K.}
$$
Interpolating only the two nearest isotherms also gives about 22 K. These are graph estimates, not exact recovered measurements.
Below $T_c$ at zero field, the stable branches have $M_0^2=-a/b$. Substitution gives the <free-energy density> relative to the omitted smooth background:
$$
F_{\min}=-\frac{a(T)^2}{4b}=-\frac{\alpha^2(T-T_c)^2}{4b}.
$$
The equilibrium excess heat-capacity <density> is $c_H=-T\partial_T^2F_{\min}=T\alpha^2/(2b)$ below $T_c$ and zero above it. Therefore
$$
\boxed{\Delta c_H=\frac{T_c\alpha^2}{2b}>0.}
$$
The all-isotherm linear fit gives \b[about $1.5\times10^3\,\mathrm{J\,m^{-3}\,K^{-1}}$]. Estimating the slope only between the 19 K and 25 K curves instead gives $\alpha\simeq6.9\times10^{-6}\,\mathrm{H\,m^{-1}\,K^{-1}}$ and a jump near $3\times10^3\,\mathrm{J\,m^{-3}\,K^{-1}}$. The lines are not exactly parallel and the intercepts are not perfectly linear in T, so \b[a jump of order a few $10^3\,\mathrm{J\,m^{-3}\,K^{-1}}$ is supported; finer precision is not]. This is per <volume> because F is a <density>; no sample <volume> is given to convert it to a total <heat capacity>.
If $b<0$, the quartic truncation is unbounded below. Add a positive sixth-order term $cM^6/6$ with $c>0$. At zero field the coexistence conditions give
$$
M_{\rm jump}^2=-\frac{3b}{4c},\qquad
a_{\rm transition}=\frac{3b^2}{16c}>0.
$$
Thus the zero-field transition is generically first order, with a discontinuous spontaneous magnetisation and possible <metastability>. It occurs before the quadratic coefficient reaches zero. The case $b=0$ is the tricritical boundary of this mean-field model; no value of c is invented.
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