= Solution
Put $\Phi(r,\theta)=R(r)Y(\theta)$ in <Laplace's equation>, multiply by $r^2/(RY)$, and separate the two variables:
$$
\frac{(r^2R')'}R=-\frac{(\sin\theta\,Y')'}{Y\sin\theta}=\lambda.
$$
With $s=\cos\theta$, the angular equation becomes $(1-s^2)Y_{ss}-2sY_s+\lambda Y=0$. Requiring a smooth <axisymmetric harmonic function> at both poles selects $\lambda=n(n+1)$ and the <Legendre polynomial> $Y=P_n(s)$, for nonnegative <integers> $n$. The radial <Cauchy-Euler equation> has solutions $r^n$ and $r^{-n-1}$. <Superposition> over the angular <eigenfunctions> therefore gives
$$
\Phi(r,\theta)=\sum_{n=0}^{\infty}(A_nr^n+B_nr^{-n-1})P_n(\cos\theta),\qquad (1-s^2)P_n''-2sP_n'+n(n+1)P_n=0.
$$
This expansion assumes regularity on each whole angular <sphere>; without it, singular angular solutions would also be possible. The boundary data in the shell involve only the $n=0,1$ modes. Define the <spherical-shell harmonic interpolation> weights
$$
p_0=\frac{a(b-r)}{r(b-a)},\quad q_0=\frac{b(r-a)}{r(b-a)},\qquad p_1=\frac{a^2(b^3-r^3)}{r^2(b^3-a^3)},\quad q_1=\frac{b^2(r^3-a^3)}{r^2(b^3-a^3)}.
$$
Each $p_n$ equals one at the inner boundary and zero at the outer one, and $q_n$ does the reverse. Hence the steady <temperature> is
$$
\boxed{T(r,\theta)=T_0p_0(r)+T_2q_0(r)+[T_1p_1(r)+T_3q_1(r)]\cos\theta.}
$$
Each term is a <harmonic function>, and substitution verifies both prescribed boundary values. Uniqueness follows from the <maximum principle for harmonic functions>.
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