= Solution
A <critical exponent> is the leading power of a singular observable as a control variable approaches a critical value. Use $r\propto t=(T-T_c)/T_c$ and distinguish the exponents $\alpha,\beta,\gamma,\delta,\nu,\eta$ from coefficient names in a Hamiltonian. Their standard definitions are
$$
M\sim(-t)^\beta,\quad \chi\sim|t|^{-\gamma},\quad C_s\sim|t|^{-\alpha},\quad
M(t=0)\sim\operatorname{sgn}(h)|h|^{1/\delta},\quad
\xi\sim|t|^{-\nu},\quad G(r;t=0)\sim r^{-(D-2+\eta)}.
$$
The <Landau approximation> predicts these powers by minimizing the <Landau free energy> and differentiating the <equation of state>. At an ordinary quartic point, $M^2=-r/u$, $\chi^{-1}=r$ above and $2|r|$ below, and $h=uM^3$ on the critical isotherm. The equilibrium singular free energy is proportional to $-r^2$ below, giving a heat-capacity jump. The quadratic fluctuation denominator is $\kappa k^2+V''$, so $\nu=1/2$ and $\eta=0$. Thus the ordinary <mean-field critical exponents> are
$$
\boxed{(\alpha,\beta,\gamma,\delta,\nu,\eta)=(0,\tfrac12,1,3,\tfrac12,0).}
$$
At a tuned tricritical point, the preceding solution gives $M\propto(-r)^{1/4}$, $V_{\min}\propto-|r|^{3/2}$ and $\chi^{-1}=4|r|$ below. At $r=0$ the <equation of state> is $h=vM^5$. The curvature still vanishes linearly, and the Gaussian gradient term still gives zero <anomalous dimension>. Therefore
$$
\boxed{(\alpha,\beta,\gamma,\delta,\nu,\eta)=(\tfrac12,\tfrac14,1,5,\tfrac12,0).}
$$
These <mean-field tricritical exponent calculations> require approach along the tuned quartic-zero direction. Fluctuations modify the powers below the appropriate <upper critical dimension>, and marginal effects can add logarithms at it.
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