= Solution
Use <inverse temperature> $\beta_T=(k_BT)^{-1}$, reserving $\beta$ without a subscript for the <critical exponent>. Write $c=qJ$; each undirected bond occurs once, so the number of bonds is $Nq/2$. In the <mean-field theory of the Ising model>, write each <Ising spin> as $\sigma_i=m+\delta\sigma_i$ and neglect the product of fluctuations:
$$
\sigma_i\sigma_j\simeq m\sigma_i+m\sigma_j-m^2.
$$
The resulting independent-spin <statistical Hamiltonian> is
$$
H_{\mathrm{MF}}=\frac{Nc}{2}m^2-(h+cm)\sum_i\sigma_i.
$$
The constant corrects the double counting of interaction energy. The <Ising spin> sums now factorize and can all be evaluated:
$$
\boxed{Z_{\mathrm{MF}}(h;m)
=e^{-\beta_TNc m^2/2}\left[2\cosh\bigl(\beta_T(h+cm)\bigr)\right]^N.}
$$
This is the approximate <partition function> at an assumed mean field; equilibrium fixes $m$ self-consistently. The corresponding <Ising auxiliary mean-field free energy> is
$$
\boxed{A_{\mathrm{aux}}(h,m)
=N\left[\frac c2m^2-k_BT\log\left(2\cosh\frac{h+cm}{k_BT}\right)\right].}
$$
Its equilibrium value gives the mean-field <Helmholtz free energy> in the imposed field. Away from a <stationary point> its parameter $m$ is an assumed field variable, not necessarily the actual mean <Ising spin> of that independent-spin distribution.
For the full small-$m$ expansion at fixed $h$, set $t_h=\tanh(\beta_Th)$ and $s_h=\operatorname{sech}^2(\beta_Th)$. Differentiating $\log\cosh z$ gives successive derivatives $t_h$, $s_h$, $-2t_hs_h$, and $2s_h(2-3s_h)$ at $z=\beta_Th$. Therefore
$$
\frac{A_{\mathrm{aux}}(h,m)}N
=-k_BT\log[2\cosh(\beta_Th)]-ct_hm
+\frac12(c-\beta_Tc^2s_h)m^2
+\frac{\beta_T^2c^3t_hs_h}{3}m^3
+\frac{\beta_T^3c^4s_h(3s_h-2)}{12}m^4+O(m^5).
$$
At zero field, <spin inversion symmetry> eliminates odd powers and this simplifies to
$$
\boxed{\frac{A_{\mathrm{aux}}(0,m)}N
=-k_BT\log2+\frac c2\left(1-\frac c{k_BT}\right)m^2
+\frac{c^4}{12(k_BT)^3}m^4+O(m^6).}
$$
The quadratic coefficient changes sign and the quartic coefficient is positive at
$$
\boxed{T_c=\frac{qJ}{k_B}=\frac{2DJ}{k_B}.}
$$
Thus the zero-field mean-field prediction is a \b[continuous, <continuous phase transition>]: the stable zero <spin magnetization> develops two symmetry-related nonzero minima continuously below $T_c$.
Both requested routes give the same <mean-field self-consistency equation>. First, the one-spin expectation in the effective field is
$$
m=\frac{\sum_{\sigma=\pm1}\sigma e^{\beta_T(h+cm)\sigma}}
{\sum_{\sigma=\pm1}e^{\beta_T(h+cm)\sigma}}
=\boxed{\tanh[\beta_T(h+cm)]}.
$$
Second, differentiating the <Ising auxiliary mean-field free energy> gives
$$
\frac1N\frac{\partial A_{\mathrm{aux}}}{\partial m}
=c\{m-\tanh[\beta_T(h+cm)]\},
$$
whose stationary condition is precisely that <mean-field self-consistency equation>. Choose its stable, lowest-free-energy branch rather than every algebraic solution.
An equally useful <mean-field approximation> parametrizes the trial distribution by its actual mean <Ising spin>. Its probabilities are $(1\pm m)/2$, giving the <Bragg-Williams free energy of the Ising model>
$$
\frac{A_{\mathrm{BW}}(h,m)}N
=-\frac c2m^2-hm+k_BT\left[
\frac{1+m}{2}\log\frac{1+m}{2}
+\frac{1-m}{2}\log\frac{1-m}{2}\right],\qquad |m|\le1.
$$
This energy-minus-entropy function has expansion
$$
\boxed{\frac{A_{\mathrm{BW}}(h,m)}N
=-k_BT\log2-hm+\frac{k_BT-c}{2}m^2+\frac{k_BT}{12}m^4+O(m^6).}
$$
Its stationary equation is $h=k_BT\operatorname{arctanh}m-cm$, again equivalent to the <mean-field self-consistency equation>. The two functions differ away from equilibrium, but agree on stationary branches: for $z=\beta_T(h+cm)$ and $m=\tanh z$, the <entropy> bracket equals $mz-\log(2\cosh z)$. Their small-$m$ coefficients therefore need not agree at arbitrary temperature; at $T_c$ they have the same leading critical quartic coefficient $c/12$. This distinction prevents confusing the auxiliary-field expansion with the physical-magnetization variational expansion.
For the <order-parameter critical exponent>, expand the equation of state at $h=0$:
$$
0=(k_BT-c)m+\frac{k_BT}{3}m^3+O(m^5).
$$
On a nonzero stable branch,
$$
m^2=\frac{3(c-k_BT)}{k_BT}+O((T_c-T)^2)
\sim\frac{3(T_c-T)}{T_c}.
$$
Hence \b[$\beta=1/2$]. The <spontaneous magnetization> is understood by selecting a branch with an infinitesimal field after the <thermodynamic limit>; a finite symmetric sample has zero exact zero-field mean <Ising spin>.
For the <magnetic susceptibility> per site, define $\chi=\partial m/\partial h$ with $h$ in energy units. Implicit differentiation gives
$$
\boxed{\chi=\frac{\beta_T(1-m^2)}{1-\beta_Tc(1-m^2)}
=\frac{1}{k_BT/(1-m^2)-c}.}
$$
Above $T_c$, $m=0$ and $\chi=1/[k_B(T-T_c)]$. Below $T_c$, evaluated on a selected ordered branch, the expansion of $m^2$ gives $\chi\sim1/[2k_B(T_c-T)]$. Thus the <magnetic-susceptibility critical exponent> is \b[$\gamma=1$ on both sides], with different amplitudes. At $T=T_c$, the equation of state becomes
$$
h=\frac c3m^3+O(m^5),\qquad
m\sim\operatorname{sgn}(h)\left(\frac{3|h|}{c}\right)^{1/3},
$$
so the <critical-isotherm exponent> is \b[$\delta=3$]. These are <mean-field critical exponents>, not a claim that neglecting fluctuations gives the exact Ising transition in every dimension.
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