(a): The key is that any $f(z)=\frac{az+b}{cz+d}$ preserves the cross ratio. It can be proved directly, but follows in a more elegant manner from the following as well:
The best if we consider $\mathbb C^2$ instead of $\mathbb C$ (the complex projective plane) and identify each $z\in\mathbb C$ to any $(az,a)$ for $a\in\mathbb C\setminus\{0\}$. If $(a,b) = (\lambda\cdot a,\lambda\cdot b)$ then they represent the same element of $\mathbb C$.
It has more benefits, for example $\infty$ can be smoothly interpreted as $(1,0)$ (represented also by any $(a,0)$).
Lemma: Assume that ${\bf u},{\bf v}\in\mathbb C^2$ are given [representing $u,v\in\mathbb C$], and ${\bf w}={\bf u}+\alpha\cdot{\bf v}$, ${\bf z}={\bf u}+\beta\cdot{\bf v}$. Then
$$(uvwz) = \displaystyle\frac\alpha\beta $$
Lemma: For $f$ as above, and if $z\in\mathbb C\cup\{\infty\}$ is represented by $\bf z$, $f(z)$ is represented by
$$\left[\begin{array}{cc} a&b\\c&d \end{array}\right]\cdot {\bf z}$$
Ok, so, suppose we know $f$ preserves the cross ratio, and that $f(z_2)=1$, $f(z_3)=0$, $f(z_4)=\infty$. Then,
$$(z_1 z_2 z_3 z_4) = (f(z_1),1,0,\infty) = \text{by def.}= \displaystyle\frac{f(z_1)-0}{1-0} = f(z_1)$$
One direction of (b) is exactly my first statement up there, and the other direction is the existence of such $f$: For given different $w_2,w_3,w_4$, it is relatively easy to construct an $f$ such that $f(w_2)=1,\ f(w_3)=0,\ f(w_4)=\infty$.
The inverse of any $f$ as above can be given by inverting the matrix (note that we can discard the constant multiplier):
$\left[\begin{array}{cc} d&-b\\-c&a \end{array}\right] $, i.e. $f^{-1}(z)=\displaystyle\frac{dz-b}{-cz+a}$.
For (c), you may need the equation of an arbitrary circle on the complex plain, say with centre $c\in\mathbb C$ and radius $\varrho>0$. Then $z$ is on this circle iff
$$|z-c|=\varrho \iff (z-c)(\bar z-\bar c) = \varrho^2 \iff \dots$$
and that being real for any $s\in\mathbb C$ means that $s=\bar s$.
Let $C$ be any circle and $R$ be the real line
[I'll use $z^*_C$ to denote the symmetric point of $z$ with respect to the $C$, and $z^* = z^*_R$ to denote the complex conjugate of $z$.]
Let $f(z) = (z,a;b,c)$ where $a,b,c$ are distinct points in $C$
Let $g(z) = (z,d;e,f)$ where $d,e,f$ are distinct points in $C$
Then $f$ is an invertible Mobius transformation that sends $(a,b,c)$ to $(1,0,\infty)$ and hence $C$ to $R$
Similarly for $g$
Let $h = g f^{-1}$
Then $h$ is an invertible Mobius transformation from $R$ to itself
Thus $h(z) = \frac{pz+q}{rz+s}$ for some $p,q,r,s \in \mathbb{R}$ because:
$\frac{p}{r} = h(\infty) \in R$ and hence WLOG $p,r \in \mathbb{R}$
$h(z) = 0$ for some $z \in R$ and hence $s \in \mathbb{R}$
$\frac{q}{s} = h(0) \in R$ and hence $q \in \mathbb{R}$
Thus $h(z^*) = \frac{pz^*+q}{rz^*+s} = \frac{p^*z^*+q^*}{r^*z^*+s^*} = (\frac{pz+q}{rz+s})^* = h(z)^*$
For any $z,z^*_C$ such that $f(z^*_C) = f(z)^*$:
$g(z^*_C) = g(f^{-1}(f(z^*_C))) = h(f(z)^*) = h(f(z))^* = g(z)^*$
Therefore the definition is independent of the three distinct points in $C$
For any Mobius transformation $T$ that maps $R$ to $C$:
Let $U = T^{-1} f^{-1}$ [You could use $U = f T$ and it would work as well.]
Then as before $U(z^*) = U(z)^*$
$T(T^{-1}(z)^*) = T(T^{-1}(f^{-1}(f(z)))^*) = T(U(f(z))^*) = T(U(f(z)^*)) = f^{-1}(f(z)^*) = z^*_C$
The uniqueness of the symmetric points was already evident from the invertibility of $f$
Best Answer
We wish to prove $$\frac{|z^*-w|\cdot|w^*-z|}{|w^*-w|\cdot|z^*-z|} = \frac{|z-\overline{w}|+|z-w|}{|z-\overline{w}|\cdot|z-w|}.$$
Note that $z^*,z,w,w^*$ is a cyclic quadrilateral, so Ptolemy implies $(1)$: $$|z^*-w|\cdot|z-w^*| = |z-z^*|\cdot|w-w^*|+|z-w|\cdot|z^*-w^*|.$$
Using this and doing a bit of algebra leads us to want to prove $$2|z-z^*|\cdot|w-w^*| = |z^*-w^*|\cdot|z-\overline{w}|-|z^*-w^*|\cdot|z-w|.$$
Now, the fact that $z^*,z,w^*,\overline{w}$ is a cyclic quadrilateral gives (via Ptolemy) $$|z^*-w^*|\cdot|z-\overline{w}| = |z-z^*|\cdot|w^*-\overline{w}|+|z^*-\overline{w}|\cdot|z-w^*|.$$
So we just wish to prove $$|z-z^*|\cdot|w-w^*| = |z^*-\overline{w}|\cdot|z-w^*|-|z^*-w^*|\cdot|z-w|.$$Now just note $|z^*-\overline{w}| = |z^*-w|$, and so we're done by $(1)$.