= Solution
Use <separation of variables>, $\Phi=F(r)H(\theta)$, in the <Laplace equation in polar coordinates>. Dividing by $FH/r^2$ gives
$$
\frac{r(rF')'}F=-\frac{H''}H=n^2.
$$
Single-valued $2\pi$-periodic angular factors have <integer> $n\geq0$. For $n\geq1$, they are $\cos n\theta,\sin n\theta$, and the radial equation $r^2F''+rF'-n^2F=0$ has solutions $r^n,r^{-n}$. The zero angular mode gives $F=a_0+b_0\log r$. These separated products give the requested solutions; arbitrary sums may use independent coefficients for each radial-angular combination.
Boundedness at the origin removes the logarithm and all negative radial powers. The boundary function is even, so its sine <Fourier coefficients> vanish. With the convention $f(\theta)=a_0+\sum_{n\geq1}a_n\cos n\theta$, its <Fourier coefficients> are
$$
a_0=\frac1{2\pi}\int_{-\pi}^{\pi}f(\theta)d\theta=\frac C2,\qquad
a_n=\frac1\pi\int_{-\pi}^{\pi}f(\theta)\cos n\theta\,d\theta=\frac{2C}{\pi n}\sin\frac{n\pi}{2}.
$$
Thus the bounded interior <harmonic extension> is
$$
\boxed{\Phi_{\rm in}(r,\theta)=\frac C2+\frac{2C}{\pi}\sum_{m=0}^\infty\frac{(-1)^m}{2m+1}\left(\frac rR\right)^{2m+1}\cos((2m+1)\theta).}
$$
For $r<R$ the series and its derivatives converge locally uniformly, so it is <harmonic>. It is a rotated and scaled instance of the <harmonic measure of a semicircle in a disk>. For example, summing the <power series> for the complex arctangent, or differentiating the following expression, gives the equivalent real form
$$
\Phi_{\rm in}=\frac C2+\frac C\pi\arctan\left(\frac{2(r/R)\cos\theta}{1-(r/R)^2}\right),\qquad r<R.
$$
This also proves boundedness directly.
\b[At the two jumps, the boundary interpretation matters.] At every continuity point the radial limit equals the prescribed value. At $\theta=\pm\pi/2$, the radial limit is $C/2$, the midpoint of the two one-sided values. Consequently for $C\ne0$ there is no harmonic solution continuous on the closed disk with exactly the printed boundary values. The standard bounded <Dirichlet problem> imposes the trace away from the jumps, or almost everywhere; assigning zero separately at those two boundary points does not alter the interior solution.
For uniqueness let $w=\widetilde\Phi-\Phi$. Formally the <divergence theorem> applied to $w\nabla w$ gives
$$
0=\int_Dw\Delta w\,dA=\int_{\partial D}w\partial_nw\,ds-\int_D|\nabla w|^2dA.
$$
For continuous zero boundary data, this immediately forces $w=0$. With the actual discontinuous data, one should justify the two exceptional points rather than assume finite <Dirichlet energy> there. Let $|w|\leq M$ and choose a cutoff $\eta_\epsilon$ which is zero within $\epsilon^2$ of those points, one beyond $\epsilon$, and linear in logarithmic distance between. Its gradient satisfies $\int|\nabla\eta_\epsilon|^2=O(1/|\log\epsilon|)$. Applying the <divergence theorem> to $\eta_\epsilon^2w\nabla w$ gives
$$
\int_D\eta_\epsilon^2|\nabla w|^2dA=-2\int_D\eta_\epsilon w\nabla\eta_\epsilon\cdot\nabla w\,dA
\leq\frac12\int_D\eta_\epsilon^2|\nabla w|^2dA+2M^2\int_D|\nabla\eta_\epsilon|^2dA.
$$
The resulting energy tends to zero. Hence $\nabla w=0$ on every compact subset and $w$ is constant; its zero boundary trace on the open arcs makes it zero. This is the <logarithmic cutoff proof of harmonic uniqueness with finitely many boundary exceptions>.
For the <exterior harmonic extension of disk boundary data>, boundedness instead excludes $r^n$ and the logarithm. The same boundary coefficients give
$$
\boxed{\Phi_{\rm out}(r,\theta)=\frac C2+\frac{2C}{\pi}\sum_{m=0}^\infty\frac{(-1)^m}{2m+1}\left(\frac Rr\right)^{2m+1}\cos((2m+1)\theta),\qquad r>R.}
$$
This is $\Phi_{\rm in}(R^2/r,\theta)$, using <inversion in a circle>. Its limit at infinity is \b[$C/2$]. To see why that value is forced independently of this construction, average any bounded exterior <harmonic function> over a circle. The mean satisfies $(r\overline\Phi')'=0$, hence is $A+B\log r$. Boundedness forces $B=0$, and the boundary mean fixes $A=C/2$. A constant limit at infinity must therefore be $C/2$.
Finally suppose two bounded exterior solutions have this limit. Their difference $w$ tends to zero. Under planar <inversion in a circle>, it is bounded and <harmonic> near the origin, so the isolated point there is removable. Its local expansion gives $w=O(r^{-1})$, $\partial_rw=O(r^{-2})$ in the original exterior. Apply the preceding <divergence theorem> energy argument on $R<r<L$, with logarithmic cutoffs at the two boundary jumps. After removing those cutoffs, the inner boundary term vanishes; the outer one is $\int_{r=L}w\partial_rw\,ds=O(L^{-2})$. Letting $L\to\infty$ forces $\nabla w=0$, and the zero limit gives $w=0$. Thus the bounded exterior solution is unique as well.
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