(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 $s=\bar s$.
Starting from scratch, let's consider the cross-ratio
$$(z_1,z_2;z_3,z_4) = \frac{z_1-z_3}{z_2-z_3}\frac{z_2-z_4}{z_1-z_4}$$
The map
$$f(z)=\frac{z_1-z_3}{z_2-z_3}\frac{z_2-z }{z_1-z }$$
sends $z_1$ to $\infty$, $z_2$ to $0$, $z_3$ to $1$, and $z_4$ to $(z_1,z_2;z_3,z_4)$. Therefore, two quadruples of points with the same cross-ratio are related by a Möbius map. The converse is also true, because cross-ratio is preserved by $z\mapsto az$, $z\mapsto z+b$, and $z\mapsto 1/z$, which generate all Möbius maps.
It remains to count the quadruples $1,-1;k,-k$ with given cross-ratio $r\in\mathbb C\setminus \{0,1\}$. We have
$$( 1,-1;k,-k) = \frac{1-k}{-1-k}\frac{-1+k}{1+k} = \left(\frac{1-k}{1+k}\right)^2$$
Here $k\mapsto \frac{1-k}{1+k}$ is a bijection from $\mathbb C\setminus \{-1,0,1\}$ onto $\mathbb C\setminus \{-1,0,1\}$. Therefore, $k\mapsto ( 1,-1;k,-k)$ is a 2-to-1 map from $\mathbb C\setminus \{-1,0,1\}$ onto $\mathbb C\setminus \{0,1\}$. Conclusion: there are precisely two suitable values of $k$ for every quadruple. (They are reciprocals of each other).
Best Answer
Let $S$ be the transformation mapping $z_1$, $z_2$, $z_3$ to $1$, $0$, $\infty$, respectively. Consider the transformation $T_k^{-1}$ mapping $1$, $0$, $\infty$ to $1$, $-1$, $k$. If we use the shorthand $z = (z_4, z_1, z_2, z_3)$, the problem reduces to finding a solution to the equation $$ T_k (-k) = z $$ as the desired transformations would then be of the form $T_k^{-1}S$.
If we plug in the expression for $T_k$, it comes down to solving the equation $$ \frac{(-k)+1}{(-k)-k}\frac{1-k}{1+1} = z $$ for $k$.