= Solution
A reciprocal vector must have integral multiples of $2\pi$ as its dot products with lattice translations. The translations along the conventional axes require $G=2\pi(h/a,k/a,l/b)$, and the body-centering translation requires $h+k+l$ to be even. Thus
$$
\boxed{\widetilde L=\{2\pi(h/a,k/a,l/b):h,k,l\in\mathbb Z,\ h+k+l\text{ even}\}.}
$$
This is a <face-centered tetragonal lattice>, with conventional dimensions $4\pi/a,4\pi/a,4\pi/b$. The <primitive unit cell> volumes are $\boxed{a^2b/2}$ in direct space and $\boxed{2(2\pi)^3/(a^2b)}$ in reciprocal space. Their conventional cell volumes are $a^2b$ and $8(2\pi)^3/(a^2b)$, containing two and four lattice points respectively.
The listed plane is $x/a+z/b=1$, whose normal reciprocal vector is $\boxed{G_0=2\pi(1/a,0,1/b)}$. It is an allowed primitive vector in that direction; half of it has nonintegral indices. The plane spacing is $d=2\pi/|G_0|=(a^{-2}+b^{-2})^{-1/2}=4a/5$ when $b=4a/3$. <Bragg's law> gives $2d\sin\theta=m\lambda$, so
$$
\boxed{\theta_m=\arcsin(5m/16),\qquad m=1,2,3.}
$$
These are approximately $18.21^\circ,38.68^\circ,69.64^\circ$. Here $\theta$ is the glancing Bragg angle; the angle between incident and scattered beams is $2\theta$.
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