= Solution
In electrostatic equilibrium, a nonzero <electric field> inside a conductor would move its free charges. Hence $E=0$ in the conducting material and the potential is constant throughout each connected conductor. Immediately outside its surface the tangential <electric field> is zero. A Gaussian pillbox across the surface gives
$$
(E_{\rm out}-E_{\rm in})\cdot n=\frac\sigma{\epsilon_0}.
$$
Since $E_{\rm in}=0$, the <electrostatic boundary conditions at a conductor> give
$$
\boxed{\sigma=\epsilon_0E_{\rm out}\cdot n}.
$$
For the widely separated shells, let their charges be $Q_1,Q_2$. Their common potential requires
$$
\frac{Q_1}{4\pi\epsilon_0R_1}
=\frac{Q_2}{4\pi\epsilon_0R_2},
\qquad Q_1+Q_2=Q.
$$
Thus the <charge sharing between distant connected spheres> is
$$
\boxed{
Q_1=\frac{R_1}{R_1+R_2}Q,
\qquad
Q_2=\frac{R_2}{R_1+R_2}Q}.
$$
For the <charge on connected concentric spherical shells>, the potentials at the two radii are
$$
\Phi(R_1)=\frac1{4\pi\epsilon_0}
\left(\frac{Q_1}{R_1}+\frac{Q_2}{R_2}\right),
\qquad
\Phi(R_2)=\frac{Q_1+Q_2}{4\pi\epsilon_0R_2}.
$$
Equality forces $Q_1=0$. Hence
$$
\boxed{Q_1=0,
\qquad Q_2=Q},
$$
so all charge lies on the exterior of the outer shell.
For the neutral sphere in the uniform field, the far-field condition gives $\alpha=-E$. Constancy of the potential on $r=R$ gives $\beta=ER^3$. Therefore
$$
\boxed{
\Phi(r,\theta)=-E\left(r-\frac{R^3}{r^2}\right)\cos\theta}.
$$
The outward normal field at the surface is
$$
E_r(R,\theta)
=-\left.\frac{\partial\Phi}{\partial r}\right|_{r=R}
=3E\cos\theta.
$$
The <induced charge on a conducting sphere in a uniform electric field> is consequently
$$
\boxed{\sigma(\theta)=3\epsilon_0E\cos\theta}.
$$
Finally,
$$
\int_{S^2}\sigma\,dA
=6\pi\epsilon_0ER^2\int_0^\pi
\cos\theta\sin\theta\,d\theta=0,
$$
confirming neutrality.
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