(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 $s=\bar s$.
Every Möbius transformation is an automorphism of the extended plane. So $Tz\in \mathbb{R}\cup\{\infty\}$ if and only if $z\in T^{-1}(\mathbb{R}\cup\{\infty\})$. Since Möbius transformations map circles (in the extended plane) to circles, that means $z$ lies on a circle passing through $z_1,z_2,z_3$. But through any three points in the extended plane, there passes only one circle, so
$$Tz\in\mathbb{R}\cup\{\infty\}\iff z \text{ lies on the unique circle passing through } z_1,z_2,z_3.$$
Best Answer
Recalling that $\text{arg}\frac{w}{z}$ is the angle between $w$ and $z$, in the first case (angles equal), $$\text{arg}\frac{z_1-z_3}{z_1-z_4}=\alpha=\text{arg}\frac{z_2-z_3}{z_2-z_4}.$$ and from there we see that $\text{arg}(z_1,z_2,z_3,z_4)=0 \implies (z_1,z_2,z_3,z_4)\in\mathbb{R}$.
The argument is similar if $z_1,z_2$ are on opposite sides of the chord.
To go in the other direction we observe that $(z_1,z_2,z_3,z_4)\in\mathbb{R}\implies \text{arg}(z_1,z_2,z_3,z_4)=0 \text{ or }\pi,$ and therefore the angles at $z_1,z_2$ subtended by the segment $z_3z_4$ are either equal or supplementary. In either case we can conclude that the points are either collinear or concyclic.
Bonus mnemonic. The cross ratio formula can be difficult to remember, whether it is for the complex plane, or for collinear points in projective geometry. This geometric interpretation gives a good mnemonic. In the diagram there are two paths to get from $z_3$ to $z_4$ via the other points: $z_3\to z_1\to z_4$, and $z_3\to z_2\to z_4$. For each path take the ratios of the path segments, and then the ratio of these ratios: $$ \frac{z_3\to z_1}{z_1\to z_4}:\frac{z_3\to z_2}{z_2\to z_4}$$
Rewriting $w\to z$ as $w-z$ we get (*) $$ \frac{z_3-z_1}{z_1-z_4}:\frac{z_3-z_2}{z_2-z_4},$$
which is equivalent to Ahlfors' definition
$$ \frac{z_1-z_3}{z_1-z_4}:\frac{z_2-z_3}{z_2-z_4}.$$
(*) Technically, instead of $w-z$ we should have rewritten $w\to z$ in 'vector' style as $z-w$ but $z-w=-(w-z)$ and all the minus signs cancel. This way we get a cross ratio formula that's easy to 'read' as two paths from $z_3$ to $z_4$.