In a possible attempt to explain a), let us focus solely on a single angle, say angle $A$. Similarly, draw tangent lines extending from the two adjacent sides, namely $AB$ and $AD$. Assuming $A\neq180$ (which we can, because it would cause $ABCD$ to be a triangle), $AB$ and $AD$ are not parallel.
This means that they meet at $A$ and continue, getting further apart as they go. If $A \lt 180$, meaning $ABCD$ is convex, $AB$ and $AD$ continue away from the shape, not intersecting any sides.
However, if $A \gt 180$, $AB$ and $AD$ enter the interior or $ABCD$ after intersecting at $A$. As the lines are infinite and the quadrilateral is not, the lines must at some point leave the shape. As two lines can only meet at a single point, and will not intersect themselves, they must leave the shape through one of the other two sides (Note Pasch's Theorem).
As both $AB$ and $AD$ are equally dependent on the angle of $A$, it is not possible for only one of the two lines to split one of the other sides.
This is evidently independent of scale and rotation. So you might as well treat the vertices as
$$
v_j = (\cos (\frac{2\pi j}{2n}), \sin (\frac{2\pi j}{2n}))
$$
or, in complex variables terms,
$$
v_j = \exp(\frac{2\pi \mathbf i j}{2n})
$$
Once you do that, your claims should all be pretty straightforward consequences of the algebra.
Best Answer
First, note that regardless of approach, we need only investigate diagonals $1$, $2$, and $3$ because there is reflective symmetry over diagonal $3$. Moreover, we only need to prove this fact for a single vertex due to rotational symmetry.
Now this probably isn't the most elegant, but one method would be to inscribe the octagon within a unit square:
$\qquad \qquad \qquad \qquad$
We can imagine this square has its lower left vertex at the origin, and we can find the coordinates of the vertices of the octagon since the lengths $x$ and $y$ obey the following:
\begin{align} &2x + y \ = \ 1 \qquad \ \quad \text{(It's a unit square)} \\ &2x^2 \ = \ y^2 \qquad \qquad \text{(From the Pythagorean theorem)} \end{align}
With those coordinates, one can explicitly compute and compare the slopes of the diagonals and of the sides.