= Solution
The <delta-shell Born amplitude> is first order in the potential strength:
$$
f_B=-gR\frac{\sin(qR)}{qR}.
$$
It retains finite-energy angular dependence and includes the leading weak-coupling contributions of all <partial waves>. By contrast, the supplied $-gR/(1+g)=-a_0$ is the \i[zero-energy limit] of the exact <S-wave scattering> amplitude. It is isotropic, depends nonlinearly on $g$, and captures the scattering-length pole missed by first-order perturbation theory.
At finite energy the exact S-wave amplitude is instead
$$
f_0(k)=\frac{e^{2i\delta_0}-1}{2ik}
=\frac1{k\cot\delta_0-ik},
$$
which is generally complex and obeys elastic unitarity. Its zero-energy limit is $-a_0$ for finite <scattering length>. Thus the two expressions printed in the question represent different approximations; the energy-independent S-wave expression is not the full finite-$k$ result of part (c).
Their common regime requires \b[both weak coupling and low energy]:
$$
\boxed{|g|\ll1,\qquad kR\ll1}.
$$
Then $qR\leq2kR\ll1$, so
$$
f_B=-gR\left[1-\frac{(qR)^2}{6}+O((qR)^4)\right],\qquad
-a_0=-\frac{gR}{1+g}=-gR+g^2R+O(g^3R).
$$
Consequently
$$
\boxed{f_B\sim f_0\sim-gR=-\frac{2mV_0R^2}{\hbar^2}}
$$
in that joint regime. Low energy alone does not remove the nonperturbative denominator $1+g$, while weak coupling alone does not remove higher-partial-wave angular dependence.
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