= Solution
For <coverage of antipodal points by a randomly oriented sector>, the house bearings differ by $\pi$. For $\phi<\pi$, no single sector can cover both. For $\phi\ge\pi$, the two allowable arcs of centre bearings overlap in two components, each of length $\phi-\pi$. Thus the <conditional probability> of a double hit is $(\phi-\pi)/\pi$, not $(\phi-\pi)/(2\pi)$. Averaging against the <probability density function> gives
$$
\mathbb P(H_1,H_2\text{ both hit})
=\frac1{\pi^3}\int_\pi^{2\pi}(\phi-\pi)(\pi-\phi/2)\,d\phi.
$$
Put $v=\phi-\pi$ to obtain $\frac1{2\pi^3}\int_0^\pi v(\pi-v)\,dv$. Therefore
$$
\boxed{\mathbb P(H_1,H_2\text{ both hit})=\frac1{12}.}
$$
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