(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
There are four permutations that don't change the value of the cross-ratio: the identity permutation and three others: \begin{align} (1\leftrightarrow2,\ 3\leftrightarrow4) \\ (1\leftrightarrow3,\ 2\leftrightarrow4) \\ (1\leftrightarrow4,\ 2\leftrightarrow3) \end{align} Since there are $24$ permutations, you should get $24/4=6$ values.
The mappings $$ \lambda\mapsto 1-\lambda,\text{ and }\lambda\mapsto \frac1\lambda, $$ generate a group of $6$ with composition of functions as the group operation, isomorphic to the group of all permutations of six elements. I'd try computing those six things in terms of $z_1,z_2,z_3,z_4$, and see if they are cross-ratios of permutations of those four.