= Solution
Let $\epsilon_B=\hbar^2g^2/(8m)$ and take $V_0>0$ as the coupling magnitude. The dispersion becomes
$$
E_\pm(q)=\epsilon_B+\frac{\hbar^2q^2}{2m}
\pm\sqrt{V_0^2+\frac{\hbar^4g^2q^2}{4m^2}}.
$$
Expanding the square root at the boundary gives
$$
E_\pm(q)=\epsilon_B\pm V_0
+\frac{\hbar^2q^2}{2m}\left(1\pm\frac{2\epsilon_B}{V_0}\right)+O(q^4).
$$
Using $1/m^*=\hbar^{-2}d^2E/dq^2$, the <band-edge effective mass in a weak periodic potential> is
$$
\boxed{\frac{m_\pm^*}{m}=\frac1{1\pm2\epsilon_B/V_0}
\simeq\pm\frac{V_0}{2\epsilon_B}\qquad(V_0\ll\epsilon_B).}
$$
The requested expression is the leading weak-potential limit. The upper band has positive <band curvature> and small positive <effective mass>; the lower band has negative curvature at its maximum and can instead be described by positive-mass <electron holes>. The small-$q$ expansion additionally requires $\hbar^2g|q|/(2m)\ll V_0$; its validity interval shrinks with the gap.
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