The first question was already answered by hardmath
using the Law of Cosines: the quadrilateral exists iff
$$
(*) \phantom{\infty\infty\infty\infty\infty\infty}
a^2 - 2ab \cos \alpha + b^2 = c^2 - 2cd \cos \beta + d^2,
\phantom{\infty\infty\infty\infty\infty\infty (*)}
$$
because both expressions must equal $e^2$. This can be used
to answer the remaining question, proving that for given $a,b,c,d$
the area of the quadrilateral is maximized when $\alpha + \beta = \pi$,
and deriving a formula for the area of the quadrilateral
in terms of $a,b,c,d$ and $\alpha,\beta$.
We first show that such a quadrilateral exists as long as there exists
any quadrilateral with sides $a,b,c,d$, that is, provided that
each of $a,b,c,d$ is smaller than the sum of the other three sides.
Indeed $\alpha + \beta = \pi$ iff $\cos \beta = - \cos \alpha$, so
(*) yields
$$
\cos \alpha = \frac{a^2+b^2-c^2-d^2}{2ab+2cd}
$$
and the necessary condition $|\cos \alpha| < 1$ becomes
$$
-(2ab+2cd) < a^2+b^2-c^2-d^2 < 2ab+2cd.
$$
If the first inequality fails then $(c-d)^2 \geq (a+b)^2$, so
$\left|c-d\right| \geq a+b$, and likewise if the second inequality fails then
$\left|a-b\right| \geq c+d$. In either case we find that $a,b,c,d$ cannot be
the sides of a quadrilateral.
Now let the area of the quadrilateral be $K$. Using Mann's
pictured decomposition of the quadrliateral into two triangles,
and the formula $\frac12 ab \sin C$ for the area of a triangle, we find
$$
2K = ab \sin \alpha + cd \sin \beta.
$$
Thus we are to maximize $2K = ab \sin \alpha + cd \sin \beta$ subject to
$$
ab \cos \alpha - cd \cos \beta = -\frac12 (a^2+b^2-c^2-d^2) =: Q.
$$
This is easily done using calculus: implicit differentiation gives
$$
ab \sin\alpha = cd \sin\beta \frac{d\beta}{d\alpha},
$$
while
$$
\frac{d(2K)}{d\alpha} = ab \cos \alpha + cd \cos \beta \frac{d\beta}{d\alpha},
$$
so if $d(2K)/d\alpha = 0$ then
$$
ab \cos\alpha = -cd \cos\beta \frac{d\beta}{d\alpha}.
$$
Dividing our two formulas for $d\beta/d\alpha$ we find
$\tan\alpha = -\tan\beta$, whence $\alpha+\beta = \pi$ as claimed.
Alternatively, we can square the formulas for $Q$ and $2K$ and add
to find
$$
Q^2 + (2K)^2
= (ab \cos \alpha - cd \cos \beta)^2 + (ab \sin \alpha + cd \sin \beta)^2
$$ $$
= (ab)^2 (\cos^2\alpha + \sin^2\alpha)
- abcd (\cos\alpha\cos\beta - \sin\alpha\sin\beta)
+ (cd)^2 (\cos^2\beta + \sin^2\beta)
$$ $$
= (ab)^2 + (cd)^2 - abcd \cos(\alpha+\beta).
$$
Thus $(2K)^2 = (ab)^2 + (cd)^2 - Q^2 - abcd \cos(\alpha+\beta)$,
which is equivalent with
Bretschneider's
formula for the area of a quadrilateral (and indeed this derivation
is equivalent to the proof recited on that Wikipedia page).
Therefore if $a,b,c,d$ are fixed then $K$ is maximized when
$\cos(\alpha+\beta) = -1$, which is to say $\alpha+\beta = \pi$.
The resulting formula for the area of a quadrilateral inscribed in a circle
is the special case of Bretschneider's formula already obtained by
Brahmagupta:
$$
K = \sqrt{(s-a)(s-b)(s-c)(s-d)},
$$
where $s = \frac12(a+b+c+d)$ is the semiperimeter. This in turn is
a generalization of Heron's formula, which is the limiting case
as one of the sides tends to zero.
