= Solution
At the <tricritical point>, $a_4=0$ and the zero-field <Landau free energy> is $f=a_2m^2+a_6m^6$. Below $T_c$, minimization gives
$$
m^4=-\frac{a_2}{3a_6}.
$$
Since $a_2\sim T-T_c$, the <order-parameter critical exponent> is
$$
\boxed{\beta=\frac14}.
$$
At this minimum, $f_{\rm sing}=a_2m^2+a_6m^6$ is proportional to $-|T-T_c|^{3/2}$. Comparing this with $f_{\rm sing}\sim|T-T_c|^{2-\alpha}$ gives the <heat-capacity critical exponent>
$$
\boxed{\alpha=\frac12}.
$$
After adding the magnetic contribution $-Bm$, the inverse zero-field <magnetic susceptibility> is the curvature $f''(m)$. Above $T_c$ it is $2a_2$, while below $T_c$ it is $-8a_2$, so $\chi\sim|T-T_c|^{-1}$ and the <magnetic-susceptibility critical exponent> is
$$
\boxed{\gamma=1}.
$$
Finally, exactly at $T_c$ the equation of state is $B=6a_6m^5$. Hence $m\sim B^{1/5}$ and the <critical-isotherm exponent> is
$$
\boxed{\delta=5}.
$$
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