= Solution
Put $t=T-T_c$, so $a_4\sim t$. At zero field the stationary equation factors as
$$
2m^3(2a_4+3a_6m^2)=0.
$$
For $t>0$, $m=0$ is the unique minimum. For $t<0$, it is unstable and the two minima are
$$
m_\pm=\pm\sqrt{-\frac{2a_4}{3a_6}}.
$$
The <order parameter> therefore tends continuously to zero, and its <order-parameter critical exponent> is $\boxed{\beta=1/2}$.
At either ordered minimum the singular free-energy density is
$$
f_{\min}=a_4m_\pm^4+a_6m_\pm^6
=\frac{4a_4^3}{27a_6^2}\sim-|t|^3,
$$
whereas it is zero for $t>0$. Two temperature derivatives give a singular <heat capacity> proportional to $|t|$ below the transition and zero above it, so the <heat-capacity critical exponent> is $\boxed{\alpha=-1}$.
The inverse <magnetic susceptibility> at a stable minimum is the curvature $\chi^{-1}=\partial^2f/\partial m^2$. Below $T_c$,
$$
\chi_-^{-1}=12a_4m_\pm^2+30a_6m_\pm^4
=\frac{16a_4^2}{3a_6},
$$
and hence $\chi_-\sim|t|^{-2}$ and $\boxed{\gamma_-=2}$. Above $T_c$, however, the curvature at $m=0$ vanishes for every $t>0$. Indeed, at small field $B=4a_4m^3+O(m^5)$, so $m\sim(B/(4a_4))^{1/3}$ and the linear susceptibility is already infinite away from the critical point. Consequently the usual <magnetic-susceptibility critical exponent> $\gamma_+$ is not defined for this exceptional free energy; assigning it a finite value would incorrectly assume a quadratic term.
At $T=T_c$, the equation of state is $B=6a_6m^5$, so $m\sim|B|^{1/5}$ and the <critical-isotherm exponent> is $\boxed{\delta=5}$. Thus the transition is continuous, although its missing quadratic term makes the high-temperature linear response singular throughout that phase.
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