= Solution
A <Bravais lattice> is the set of all integer combinations of linearly independent primitive translation vectors. Its <reciprocal lattice> consists of $g$ satisfying $g\cdot l\in2\pi\mathbb Z$ for every translation $l$; equivalently $e^{ig\cdot l}=1$.
The original PDF's third family has a bare $j$, not $n_2j$. Taken literally, the listed set is not the claimed <Bravais lattice>: the claimed primitive vector $a(i+k)/2$ does not occur. Its half-integer first and third coordinates force it into that third family, whose second coordinate is always $a$, not zero. The intended repair is \b[$j\mapsto n_2j$ in the third family]. The following conclusions apply to that corrected set.
The four corrected families are exactly
$$
L=\frac a2\{(h,k,l)\in\mathbb Z^3:h+k+l\text{ is even}\}.
$$
The allowed parity patterns are all even, or exactly two odd coordinates. An integer combination of the proposed primitive vectors has coordinates $(a/2)(m_2+m_3,m_1+m_3,m_1+m_2)$, so lies in this set. Conversely, solve
$$
m_1=\tfrac12(k+l-h),\quad m_2=\tfrac12(h+l-k),\quad m_3=\tfrac12(h+k-l).
$$
All three are integers precisely when $h+k+l$ is even. This proves that the corrected set is the <face-centered cubic lattice> with the proposed primitive basis.
Direct dot products give $a_i\cdot b_j=2\pi\delta_{ij}$ for
$$
\boxed{b_1=\frac{2\pi}a(-1,1,1),\quad b_2=\frac{2\pi}a(1,-1,1),\quad b_3=\frac{2\pi}a(1,1,-1).}
$$
These vectors generate the entire <reciprocal lattice>, since the coefficients of an arbitrary reciprocal vector in this basis are its integer pairings with the $a_i$. Its vectors are $(2\pi/a)(h,k,l)$ with all three coordinates of the same parity, the <body-centered cubic lattice>.
In <Bragg scattering>, translating a scattering atom by $l$ changes the amplitude by $e^{i(k-k')\cdot l}$. Constructive interference from every lattice translate requires this factor to be one, so $k'-k=g$ is a <reciprocal lattice> vector. For elastic scattering, $|k'|=|k|$; the chord between these two vectors has length $2|k|\sin(\theta/2)$. Therefore
$$
\boxed{\sin(\theta/2)=\frac{|g|}{2|k|}.}
$$
The shortest nonzero reciprocal vectors are the $(111)$ family of length $(2\pi/a)\sqrt3$, followed by the $(200)$ family of length $(2\pi/a)2$. Identical atoms on this <Bravais lattice> introduce no additional basis extinction. If both families are accessible at the incident wavelength, their scattering angles obey \b[$\boxed{\sin(\theta_1/2)/\sin(\theta_2/2)=\sqrt3/2}$].
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