= Solution
Use reduced variables $t=T/T_c$, $v=V/V_c$ and $P=p/p_c$, so $P=8t/(3v-1)-3/v^2$. Write $t=1-\tau$, $v=1+s$. Near the <critical point>,
$$
P=1-4\tau+6\tau s-\frac32s^3+O(\tau s^2+s^4).
$$
Set $s=\sqrt\tau\,z$. The leading departure from the coexistence pressure is $\tau^{3/2}(6z-3z^3/2)$. Equal pressure at the two outer endpoints together with the Maxwell equal-area condition selects the symmetric pair $z_l=-2$, $z_g=2$, with coexistence pressure $1-4\tau+O(\tau^2)$. To see why, a horizontal line through the odd cubic has zero signed area between its outer equal-height roots only at zero height: the signed area is strictly decreasing with line height, and symmetry makes it zero there. Thus
$$
\boxed{V_g-V_l=4V_c\sqrt{\frac{T_c-T}{T_c}}+O(T_c-T).}
$$
This proves the coexistence exponent one half.
Along the critical isotherm $T=T_c$ itself, the expansion instead gives
$$
\boxed{p-p_c=-\frac{3p_c}{2V_c^3}(V-V_c)^3+O((V-V_c)^4).}
$$
The cubic scaling therefore has exponent three, with a negative coefficient. A fixed noncritical isotherm does not pass through the <critical point>, so that qualification is essential.
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