= Solution
The central <point charge> and spherical boundaries give a <spherically symmetric charge distribution>. A <conductor in electrostatic equilibrium> has zero <electric field> within its material. Consequently a spherical <Gaussian surface> inside the metal encloses zero net <electric charge>, so the inner surface carries $-q$. The shell's prescribed total <electric charge> then leaves $-2q$ on its outer surface. <Gauss's law> gives
$$
\boxed{\mathbf E(r)=
\begin{cases}
\dfrac{q}{4\pi\epsilon_0r^2}\,\hat{\mathbf r},&0<r<r_1,\\[4pt]
\mathbf0,&r_1<r<r_2,\\[4pt]
-\dfrac{2q}{4\pi\epsilon_0r^2}\,\hat{\mathbf r},&r>r_2.
\end{cases}}
$$
The <electric field> is singular at the ideal central <point charge>. At each charged surface the normal component has distinct one-sided limits, as required by the <electrostatic boundary conditions at a conductor>. Spherical symmetry makes the outer <surface charge density> uniform:
$$
\boxed{Q_{\rm inner}=-q,\qquad \sigma_{\rm outer}=-\frac{2q}{4\pi r_2^2}.}
$$
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