(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 $s=\bar s$.
You can show that if the cross ratio of $(ABCD)$ is $m$, then the cross ratio of $(ACBD)$ is $1-m$. This is independent of where the points are, and follows directly from writing it out.
You mentioned that $ABCD$ is concyclic if and only if the cross ratio is real. There is an additional fact that if $AB$ does not separate $CD$, then the cross ratio is positive.
Hence, we have $0 < m < 1$. Applying absolute value, Ptolemy's Theorem follows.
Best Answer
The cross ratio, as a function of the first argument, is a Möbius transformation,
$$T \colon z \mapsto (z,z_2,z_3,z_4).$$
We have $T(z_2) = 0$, $T(z_3) = 1$, $T(z_4) = \infty$ (or some permutation thereof, depends on the used definition of the cross ratio). So if we know that Möbius transforms map circles or straight lines to circles or straight lines, we know that $T$ maps the circle or straight line passing through the three points $z_2,\,z_3,\,z_4$ to the real line (plus infinity). Since Möbius transformations are bijective, $T(z)$ is real (or $\infty$) if and only if $z = T^{-1}(Tz)$ lies on the circle or straight line passing through $z_2,\,z_3,\,z_4$, which is the image of the real line (plus $\infty$) under $T^{-1}$.