= Solution
For the nonconstant boundary data, use <separation of variables> in <polar coordinates> and put $\phi=R(r)\cos2\theta$. <Laplace's equation> reduces to
$$
r^2R''+rR'-4R=0.
$$
This <Cauchy-Euler equation> has independent solutions $r^2$ and $r^{-2}$. Unlike a disk containing the origin, the <annulus> permits both. The zero inner boundary value requires $R=A(r^2-a^4r^{-2})$, and the outer value requires $A=Vb^2/(b^4-a^4)$. Hence
$$
\boxed{\phi_0(r,\theta)=V\frac{b^2(r^4-a^4)}{r^2(b^4-a^4)}\cos2\theta,\qquad a<r<b.}
$$
Substituting $r=a$ gives zero, while $r=b$ gives $V\cos2\theta$. This is the <Dirichlet problem on an annulus for one Fourier mode>; no other angular mode is needed.
For the changed inner boundary value, use <superposition> to add a radial <harmonic function> that is $-V$ at $a$ and zero at $b$. The radial zero-mode equation is $(rR')'=0$, giving $R=C+D\log r$. Imposing the two constant boundary values gives
$$
R(r)=-V\frac{\log(b/r)}{\log(b/a)}.
$$
Therefore the second answer is
$$
\boxed{\phi(r,\theta)=V\frac{b^2(r^4-a^4)}{r^2(b^4-a^4)}\cos2\theta-V\frac{\log(b/r)}{\log(b/a)}.}
$$
The logarithmic term is independent of $\theta$, so it changes only the angular average of the <electric potential>. Both answers are unique for these <Dirichlet boundary conditions>, by uniqueness for <Laplace's equation>.
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