= Solution
A three-dimensional <Bravais lattice> is the set of all integer combinations of three linearly independent primitive vectors. Its <reciprocal lattice> is
$$
\Lambda^*=\{G:G\mathbin{\cdot}R\in2\pi\mathbb Z
\text{ for every }R\in\Lambda\},
$$
and the first <Brillouin zone> is the <Wigner-Seitz cell> of the origin in reciprocal space.
For the stated <FCC lattice>, the primitive-cell volume is
$$
\Omega=a_1\mathbin{\cdot}(a_2\times a_3)=\frac{a^3}{4}.
$$
The reciprocal-basis formula gives
$$
\boxed{
b_1=\frac{2\pi}{a}(-e_1+e_2+e_3),\quad
b_2=\frac{2\pi}{a}(e_1-e_2+e_3),\quad
b_3=\frac{2\pi}{a}(e_1+e_2-e_3)}.
$$
A reciprocal primitive cell, and hence the <Brillouin zone>, has volume
$$
\boxed{\frac{(2\pi)^3}{\Omega}=\frac{32\pi^3}{a^3}}.
$$
In the first <Born approximation>, with momentum transfer $q=k-k'$, the <scattering amplitude> is
$$
f_V(k;k')=-\frac{m}{2\pi\hbar^2}
\int_{\mathbb R^3}e^{iq\cdot r}V(r)\,d^3r.
$$
For the spherical shell, align the polar axis with $q$ and use its radial <Dirac delta function>:
$$
\int e^{iq\cdot r}V_0\delta(r-d)\,d^3r
=4\pi V_0d^2\frac{\sin(qd)}{qd}.
$$
Therefore
$$
\boxed{\widetilde f_V(k;k')
=-\frac{2mV_0d^2}{\hbar^2}\frac{\sin(qd)}{qd}}.
$$
Translation of each atomic potential multiplies its transform by a phase, so the <crystal lattice structure factor> gives
$$
f_{V_\Lambda}(k;k')
=\widetilde f_V(k;k')\sum_{R\in\Lambda}e^{iq\cdot R}.
$$
For an infinite lattice, the lattice <Poisson summation formula> turns the sum into
$$
\sum_{R\in\Lambda}e^{iq\cdot R}
=\frac{(2\pi)^3}{\Omega}
\sum_{G\in\Lambda^*}\delta^{(3)}(q-G).
$$
Thus the amplitude vanishes away from reciprocal-lattice momentum transfers.
In <elastic scattering>, $|k|=|k'|$ and hence
$$
|q|=2k\sin\frac\theta2.
$$
The two shortest relevant reciprocal-vector lengths of this <FCC lattice> are
$$
|G_2|=\frac{2\pi\sqrt3}{a},
\qquad |G_1|=\frac{4\pi}{a}.
$$
Both are kinematically available when $k>2\pi/a$. Assigning their scattering angles $\theta_2$ and $\theta_1$, respectively, gives
$$
\boxed{\frac{\sin(\theta_1/2)}{\sin(\theta_2/2)}
=\frac{|G_1|}{|G_2|}=\frac2{\sqrt3}}.
$$
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