= Solution
Use <separation of variables>, $\phi=R(r)\Theta(\theta)$. After multiplying the <Laplace equation> by $r^2/(R\Theta)$, the two separated equations are
$$
\Theta''+\lambda\Theta=0,\qquad r^2R''+rR'-\lambda R=0.
$$
Single-valuedness requires $2\pi$-periodicity. Thus the nonconstant angular modes have $\lambda=n^2$, $n=1,2,\ldots$, and angular factors $\cos n\theta,\sin n\theta$. Their radial factors are $r^n,r^{-n}$. The zero mode has constant angular factor and radial solution $A+B\log r$; a term proportional to $\theta$ would be multivalued.
Finiteness at the origin removes the logarithm and negative powers. In the <disk> the general regular harmonic expansion is
$$
\boxed{\phi=A_0+\sum_{n\ge1}r^n(A_n\cos n\theta+B_n\sin n\theta)}.
$$
For the exterior, regularity at every finite radius allows
$$
\boxed{\phi=C_0+D_0\log r+\sum_{n\ge1}\big[(C_nr^n+D_nr^{-n})\cos n\theta+(E_nr^n+F_nr^{-n})\sin n\theta\big]}.
$$
Coefficients must give convergent harmonic expansions on the domain and the intended boundary regularity at $r=a$. If “finite” is intended to mean bounded as $r\to\infty$, then $D_0=C_n=E_n=0$ and only the constant and decaying modes remain. Infinity is not included in the printed interval; moreover part (b)'s uniform background flow necessarily has an unbounded potential. Stating both interpretations avoids discarding that background mode.
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