= Solution
Within each constant-conductivity region the <temperature> is <harmonic>. For $u=v(r)\cos\theta$, the spherical Laplacian gives
$$
\Delta u=\left(v''+\frac2rv'-\frac2{r^2}v\right)\cos\theta.
$$
Trying $v=r^s$ yields $s(s-1)+2s-2=(s-1)(s+2)=0$. Thus $v=Cr+Dr^{-2}$. Boundedness at the origin removes the interior $r^{-2}$ term, while the prescribed far <gradient> fixes the exterior growing term:
$$
u_{\mathrm{in}}=C_{\mathrm{in}}r\cos\theta,\qquad
u_{\mathrm{out}}=(r+Br^{-2})\cos\theta.
$$
Continuity of <temperature> and normal flux at $r=r_1$ give
$$
C_{\mathrm{in}}=1+\frac{B}{r_1^3},\qquad
a_1C_{\mathrm{in}}=a_0\left(1-\frac{2B}{r_1^3}\right).
$$
Solving these two equations yields $B=-r_1^3(a_1-a_0)/(a_1+2a_0)$ and $C_{\mathrm{in}}=3a_0/(a_1+2a_0)$. Consequently
$$
\boxed{
u_{\mathrm{out}}=x_3-\frac{r_1^3(a_1-a_0)}{a_1+2a_0}\frac{x_3}{|x|^3},
\qquad
u_{\mathrm{in}}=\frac{3a_0}{a_1+2a_0}x_3.
}
$$
Positive conductivities make the denominator nonzero. The matched solution is unique in the usual finite-energy decaying-perturbation class: the difference of two solutions has zero far data and interface jumps, and an energy integration gives zero <gradient>.
For the <polarization field of a conductivity inclusion>, use the sign convention consistent with the perturbation equation:
$$
P=(a-a_0I)\nabla u,\qquad
\beta=\frac{3a_0(a_1-a_0)}{a_1+2a_0}.
$$
Thus
$$
\boxed{P(x)=\beta\,\chi_B(x)e_3.}
$$
For a general incident constant <gradient> $E$, rotational <symmetry> and <linearity> replace $e_3$ by $E$. The polarization vanishes outside the <sphere> because its local conductivity equals the reference value. The <temperature> perturbation outside is dipolar; this is the <spherical conductivity inclusion> response used in the remaining parts.
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