Determine an hyperbolic midpoint I am asked to determine the hyperbolic midpoint of the points $0,\frac{1}{2} \in \mathcal{P}$
Q: how do I determine the hyperbolic midpoint and what is actually meant by the midpoint? 
 A: Poincaré model
I'm not sure what $\mathcal P$ is in your notation. The Poincaré disk, as a subset of $\mathbb C$?
If you restrict yourself to the Poincaré disk, then the midpoint $M$ of the geodesic segment $AB$ would be a point on the line $AB$ which has equal and finite hyperbolic distance from both $A$ and $B$.
In your specific case, the line $AB$ would be the real axis. For distances you can refer to e.g. Wikipedia:

If $u$ and $v$ are two vectors in real $n$-dimensional vector space $\mathbf R^n$ with the usual Euclidean norm, both of which have norm less than $1$, then we may define an isometric invariant by $$\delta (u, v) = 2 \frac{\lVert u-v \rVert^2}{(1-\lVert u \rVert^2)(1-\lVert v \rVert^2)}$$ where $\lVert \cdot \rVert$ denotes the usual Euclidean norm. Then the distance function is $$d(u, v) = \operatorname{arcosh} (1+\delta (u,v))$$

Actually the first half of that statement is enough: you don't need the numeric values of distances, you just need equality of distances, so you can aim for $\delta(A,M)=\delta(M,B)$ or more specifically $\delta(0,M)=\delta(M,\frac12)$.
\begin{align*}
2\frac{M^2}{1-M^2} &=
2\frac{(M-\frac12)^2}{(1-M^2)\frac34}
\\
3M^2(1-M^2) &=
4(M-\tfrac12)^2(1-M^2)
\\
M^4 - 4M^3 + 4M - 1 &= 0
\\
(M - 1)(M + 1)(M^2 - 4M + 1) &= 0
\end{align*}
So there are four solutions algebraically. Two are ideal points, which have the same distance to $A$ and $B$ since they have infinite distance. The other two are the solutions of the quadratic equation, namely $M=2\pm\sqrt3$. Of these, only one lies within the Poincaré disk, so that's the one to choose. The other is its inverse, because algebraically speaking, points which are related via inversion in the unit circle are essentially the same in the Poincaré model.
Cayley-Klein metric
When you model the hyperbolic plane using the Beltrami-Klein model, or in different words using a Cayley-Klein metric on the real projective plane, then you'd end up with two distinct midpoints. One of them would be the same as above, while the other one would have equal but complex distance from both endpoints. One of them lies outside the fundamental conic and therefore has no corresponding point in the Poincaré model. So what you consider a midpoint, and how many there are, depends a lot on your model, and which points you include in your considerations.
