= Solution
Place eight identical cubes around the charged corner to make one larger cube, with the <point charge> at its centre. Each large face has <electric flux> $q/(6\epsilon_0)$ and consists of four equivalent small faces. Each of the original cube's three faces opposite the charged corner therefore has
$$
\boxed{\Phi_{\rm opposite}=\frac{q}{24\epsilon_0}.}
$$
On each of its three faces containing the corner, the <electric field> of the <point charge> obeying <Coulomb's law> is tangent to the face at every regular point; its outward-normal component is zero. Defining the face integral by omitting the singular corner gives
$$
\boxed{\Phi_{\rm incident}=0,\qquad\Phi_{\rm original\ cube}=\frac{q}{8\epsilon_0}.}
$$
A boundary <point charge> does not satisfy the ordinary nonsingular hypotheses for simply counting it as an interior charge in <Gauss's law>. The $1/8$ result expresses the octant <solid angle> and the specified improper surface integrals. Moving a charge towards the corner from inside a cube before integrating is a different limiting operation; its total closed-surface <electric flux> stays $q/\epsilon_0$ until it reaches the boundary.
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