Thank you to Roy Madar from Allerton High School and Andrei Lazanu from Tudor Vianu National College, Bucharest, Romania for their solutions.

Draw two concentric circles with centres at the origin and radii $|z|$ and $|w|$ and suppose, without loss of generality, that $|z|> |w|$ then $z$ lies somewhere on the outer circle and $w$ lies somewhere on the inner circle.

In this representation, working with complex numbers is
similar to working with vectors. The two complex numbers can be
added and subtracted as vectors. So, to subtract them, it is the
same as adding one to the opposite of the other. The result is
represented in the figure. The distance between the two points
representing the complex numbers $z$ and $w$ is the modulus of
$z-w$ written as $|z-w|$. The lengths of the sides of the triangle
in the diagram are $|z|$, $|w|$ and $|z-w|$.

The shortest distance between any two points where one is on
each circle is given by the difference of the radii of the two
circles $|z|-|w|$. This would be the distance between the points
$z$ and $w$ if they had the same argument. If $z$ and $w$ have
different arguments then the distance between the points will be
$|z- w|$ and hence $|z-w|\geq |z| - |w|$.

For the second part, mark points $A$ and $B$ on the unit
circle to represent $e^{i\alpha*}$ and $e^{i\beta*}$ respectively
where $\alpha*$ and $\beta*$ lie between 0 and $2\pi$. To include
all possible values of $\alpha$ and $\beta$ we have $\alpha
=\alpha* +2m\pi$ and $\beta = \beta* + 2n\pi$ for some integers $m$
and $n$. Clearly $A$ and $B$ also represent $e^{i\alpha}$ and
$e^{i\beta}.$

The chord length $AB$ is less than or equal to the arc length
$AB$. Hence $$|e^{i\alpha}-e^{i\beta}|\leq \alpha - \beta.$$