Solving Dirichlet problem on exterior of unit circle 
Let $C$ be the unit circle $|z|=1$ and $h$ a continuous real function on $C$. Find a function $u$ which is harmonic in $\mathbb{C} \setminus \mathbb{D}_1(0)$ with boundary condition $u(z) = h(z)$ for $z \in C$. ($u(z)$ is short-hand for $u(x,y)$ where $z = x+iy$).
i.e. solve the Dirichlet problem on the exterior of the unit disk.

I know how to solve such a problem using PDE's and Fourier series, but my aim is to do this problem using complex analysis (particularly, I am trying to use the solution I already have for the Dirichlet problem in the unit disk coupled with the properties of Moebius maps).
The solution to the Dirichlet problem in the unit disk with boundary condition $h$ is:
$$u(re^{i \phi}) = \frac{1-r^2}{2 \pi} \int_0^{2 \pi} \frac{h(e^{i \theta})}{1+r^2-2r \cos(\phi-\theta)} d\theta$$
where $0 \le r < 1$ and $0 \le \phi \le 2 \pi$
Now, $w = \frac1{z}$ is a bijective conformal map form the unit disk to the exterior of the unit disk. Moreover, it is its own inverse, so it also takes the exterior of the unit disk to the unit disk.
Then $h^\ast (w) = h(z)$ is continuous on the circle $|w| = 1$. Let $u^\ast(w)$ be the unique solution of the Dirichlet problem on the unit disk $|w| < 1$ with boundary values $h^\ast(w)$.
Let $u(z) = u^\ast(w(z))$. Then $u$ is harmonic in $\mathbb{C} \setminus \mathbb{D}_1(0)$ and takes on the desired boundary values $h(z)$.
Would it now be correct to simply say, for $0 \le r' < 1$, $0 \le \phi' \le 2 \pi$:
$$u^\ast(r'e^{i \phi'}) = \frac{1-r'^2}{2 \pi} \int_0^{2 \pi} \frac{h^\ast(e^{i \theta})}{1+r'^2-2r' \cos(\phi'-\theta)} d\theta$$
implies
$$u(\frac1{r'}e^{-i \phi'}) = \frac{1-r'^2}{2 \pi} \int_0^{2 \pi} \frac{h(e^{-i \theta})}{1+r'^2-2r' \cos(\phi'-\theta)} d\theta$$
Letting $r = \frac1{r'}$ and $\phi = -\phi'$, we get
$$u(re^{i \phi}) = \frac{1-\frac1{r^2}}{2 \pi} \int_0^{2 \pi} \frac{h(e^{-i \theta})}{1+\frac1{r^2}-\frac2{r} \cos(\phi+\theta)} d\theta$$
$$u(re^{i \phi}) = \frac{r^2-1}{2 \pi} \int_0^{2 \pi} \frac{h(e^{-i \theta})}{1+r^2-2r \cos(\phi+\theta)} d\theta$$
For $r > 1$ and $0 \le \phi \le 2 \pi$
 A: Yes, your computation is correct. It would be a little easier if it stayed more in the spirit of complex analysis: the Poisson kernel for the unit disk is 
$$
P(z, \zeta) = \frac{1}{2\pi}\frac{1-|z|^2}{|z-\zeta|^2}
$$
Plugging $1/\bar z$ for $z$ we get
$$
P(1/\bar z, \zeta) = \frac{1}{2\pi}\frac{1-1/|z|^2}{|1/\bar z-\zeta|^2}
= \frac{1}{2\pi}\frac{|z|^2-1}{|1-\bar z\zeta|^2} 
= \frac{1}{2\pi}\frac{|z|^2-1}{| \bar z - \bar \zeta |^2}
= \frac{1}{2\pi}\frac{|z|^2-1}{|z - \zeta|^2}
$$
Since the transformation $z\mapsto 1/\bar z$ is anticonformal, it preserves harmonic functions (same proof as for conformal maps; or just think of it as composition of a conformal map with conjugation; the conjugation preserves harmonic functions). Also, it is the identity on the boundary of the unit disk. Conclusion: for $|z|>1$, 
$$
u(z) = \frac{1}{2\pi}\int_{|\zeta|=1}\frac{|z|^2-1}{|z - \zeta|^2} \, h(\zeta)\,|d\theta|
$$
This is essentially the same that you got, but using $1/\bar z$ is a little simpler: in polar coordinates, it maps $re^{i\theta}$ to $r^{-1}e^{i\theta}$, without flipping the sign of polar angle. 
In your notation, un-flipping the sign of $\theta$ makes the final formula nicer:
$$u(re^{i \phi}) = \frac{r^2-1}{2 \pi} \int_0^{2 \pi} \frac{h(e^{i \theta})}{1+r^2-2r \cos(\phi-\theta)} d\theta$$
