Find point on ellipse arc Known ellipse semiaxis, points F and G, angles alpha and beta. How to find point H?

 A: If you draw the line $FG$, and draw the angle $\alpha$ at $F$, then you can draw the axes and essentially have the ellipse. So the interesting part is finding a match with $\beta$.
Let's look at this problem algebraically for a moment. If $H=(x,y)$, then one can compute the normal of the tangent direction as a linear function in these variables. So assume that normal to be $(ax+by+c,bx+dy+e)$. The coefficients $a$ through $e$ can be read from the matrix of the conic. If you have $G=(g,h)$, then the direction between $G$ and $H$ is $(x-g,y-h)$ and the normal is $(y-h,g-x)$. Since the cosine is proportional to the dot product you get
$$\cos^2\beta=\frac{\bigl((y-h)(ax+by+c) + (g-x)(bx+dy+e)\bigr)^2}
{\bigl((y-h)^2+(g-x)^2\bigr)\bigl((ax+by+c)^2+(bx+dy+e)^2\bigr)}$$
If you move the denominator to the other side, and expand both sides, you will end up with an algebraic curve of degree 4. Intersecting that with your original conic will be no fun at all. You can expect up to 8 points of intersection, i.e. 8 different valid solutions. I very much doubt you'll find a geometric construction for this.
I might of course be wrong. This algebraic curve of degree 4 might factor into simpler curves for some reason, making the whole problem easier than it looks now. So I'd suggest you start with some generic example, work through that algebraically and try to see whether indeed there are 8 distinct solutions. If so, I'd give up looking for constructions at that point, but you may of course decide to continue nonetheless.
In my numeric experiments, four of the assumed eight solutions were always complex, the other four real. For example, the ellipse $x^2+4y^2=1$ with $G=(1,1)$ and $H_1=(0.6,0.4)$ looks like this:

Sure, some of the $H_i$ have the angle in the opposite direction (since the cosine doesn't care about the sign, and the squaring would throw away the sign even if we were using the sine instead), but I can't think of a geometric construction that would allow you to take that sign into account a priori, instead of simply as a means to exclude some solutions a posteriori. And if computed using algebraic numbers, the results depicted have minimal polynomials of degree up to 4. So it might be that there is a clever way to throw away the complex solutions up front, and only compute the real ones, but that still leaves you with four solutions and a hard time finding a construction sequence for them.

Edit: I've created some randomized setups, and found several where there are indeed 8 distinct real solutions, e.g. like this:

The above illustration is for $\lvert\beta\rvert=60°$, the conic $x^2+9y^2=3$ and the point $G=(0,-1/2)$. So I no longer think it might be possible to simplify this to something which yields only four solutions. Furthermore, since some of the minimal polynomials of the resulting coordinates have degree 3, I know that they are not constructible numbers.
So a construction of these solutions using compass and straightedge is impossible!
I suggest you use algebraic or numeric methods if faced with a problem like this. Or perhaps some form of iterative refinement if you actually have to create this on paper.
