I too would interpret the symbols $\alpha$, $\beta$, and $\gamma$ as angles
between the sides, and $a$ as the arc length of a side.
I know of two sets of formulas that might be called the spherical law of cosines.
You can find both of them on the page for
spherical trigonometry at Wolfram MathWorld.
One set of formulas they call the cosine rule for sides:
$$\cos a = \cos b \cos c + \sin b \sin c \cos\alpha$$
(with corresponding formulas with $\cos b$ and $\cos c$, respectively, on the left).
The other set of formulas they call the cosine rule for angles:
$$\cos\alpha = -\cos\beta \cos\gamma + \sin\beta \sin\gamma \cos a$$
(with corresponding formulas with $\cos\beta$ and $\cos\gamma$, respectively, on the left).
You can easily put any of these formulas in a format like the one you have used;
from the cosine rule for angles you would end up with
$$\cos a = \frac{\cos\alpha + \cos\beta \cos\gamma}{\sin\beta \sin\gamma}.$$
Note the "$+$" operation in this formula where you have "$-$".
It turns out this has no effect on your result, since the right-hand operand
turns out to be zero.
Where you make a misstep is here:
$$\frac{\sqrt{6}+\sqrt{2}}{2}=4\cos(\alpha)$$
$$2(\sqrt{6}+\sqrt{2})=\cos(\alpha)$$
You need to divide both sides of the first equation by $4$, so the second equation
should have been
$$\frac{\sqrt{6}+\sqrt{2}}{8} = \cos\alpha.$$
One way you might notice the error is that $2(\sqrt{6}+\sqrt{2}) > 1$,
meaning it cannot be the cosine of any real-valued angle.
The result is not a "nice" angle like $60$ degrees or $75$ degrees, so
either you give a calculator's approximation or you leave it as an expression
involving $\cos^{-1}$.
You could find the other two sides of the triangle using law of cosines,
or you could use the Law of Sines for a spherical triangle, which is
relatively easy to remember:
$$\frac{\sin\alpha}{\sin a} = \frac{\sin\beta}{\sin b} = \frac{\sin\gamma}{\sin c}.$$
That's actually three equations, two of which allow you to solve for your
unknown sides $b$ and $c$.
Best Answer
From the polar/dual law of cosines applied to the upper triangle,
$$\cos\varphi=-\cos(\pi-\alpha)\cos(\pi-\beta)+\sin(\pi-\alpha)\sin(\pi-\beta)\cos AB,$$
$\varphi$ determines the edge length $AB$ and vice-versa, assuming the angles $\alpha,\beta$ are given. So your question is whether the edge lengths can be determined from $\alpha,\beta,\gamma,\delta$.
(Equivalently by duality, your question is whether a spherical quadrilateral's angles can be determined from its edge lengths.)
The answer is no. Consider a "rectangle", defined by symmetry rather than right angles, so that $\alpha=\beta=\gamma=\delta$, $AB=CD$, $AD=BC$. Put the rectangle's centre at $(0,0,1)\in\mathbb S^2\subset\mathbb R^3$, and define the edges' outward normal vectors (tangent to the sphere):
$$e=(\cos\lambda,\;0,\;-\sin\lambda)$$
$$f=(0,\;\cos\mu,\;-\sin\mu)$$
$$g=(-\cos\lambda,\;0,\;-\sin\lambda)$$
$$h=(0,\;-\cos\mu,\;-\sin\mu).$$
We require
$$\cos(\pi-\alpha)=e\cdot f=f\cdot g=g\cdot h=h\cdot e$$
$$=\sin\lambda\sin\mu$$
to be constant, so either of $\lambda,\mu$ determines the other. We still have $1$ degree of freedom.
A vertex vector is perpendicular to the two edges' normal vectors; one of them is
$$\pm\frac{e\times f}{\lVert e\times f\rVert}=\pm\frac{(\sin\lambda\cos\mu,\;\cos\lambda\sin\mu,\;\cos\lambda\cos\mu)}{\sqrt{1-\sin^2\lambda\sin^2\mu}}$$
and an adjacent one is
$$\pm\frac{f\times g}{\lVert f\times g\rVert}=\pm\frac{(-\sin\lambda\cos\mu,\;\cos\lambda\sin\mu,\;\cos\lambda\cos\mu)}{\sqrt{1-\sin^2\lambda\sin^2\mu}}$$
so the cosine of the distance between these two vertices is
$$\pm\frac{e\times f}{\lVert e\times f\rVert}\cdot\frac{f\times g}{\lVert f\times g\rVert}=\pm\frac{1-2\sin^2\lambda+\sin^2\lambda\sin^2\mu}{1-\sin^2\lambda\sin^2\mu}$$
$$=\pm\frac{1+\cos^2\alpha-2\sin^2\lambda}{\sin^2\alpha}$$
which is not constant. In other words, the edge length is not determined by the angle $\alpha$.