If $E\supset F\supset k$ then let $K$ be a finite solvable Galois extension of $k$ containing F and $L$ be a solvable Galois extension of $K$ containing $EK$, If $\sigma$ is any embedding of $L$ over $k$ then in a given algebraic closure then $\sigma K=K$ and hence $\sigma L$ is a solvable extension of K
Questions:
$\bullet$ Is it the same to be finite solvable or solvable (because a solvable extension must be finite I assume)
$\bullet$ $\sigma K=K$ follows from that $\sigma$ is k-linear and the $K$ is finite over $k$ hence any injective map is also surjective (correct me if I'm wrong)
$\bullet$ How do you conclude that $\sigma L$ is solvable (is it not Galois anymore ?)
Let's see if I can clear out some things here:
1) A "solvable extension" is just one with a solvable automorphism group. Finite means the extensions is a finite dimensional one. Both things don't mean the same although it could be some texts assume all the time, or in some specific chapters, that all extensions are finite, just as in many basic linear algebra texts finite dimensionality is many times a basic assumption.
2) The equality $\;\sigma K=K\;$ follows from the fact that, if $\;\sigma L\to\overline k\;$ is the embedding we began with to some algebraically closed field $\;\overline k\;$ containing $\;k\;$ , then the restriction $\;\sigma|_K:K\to\overline k\;$ is an embedding of $\;K\;$ into the same alg. closed field, and since $\;K/k\;$ is normal (as it is Galois) it follows this embedding must be an element of Aut$\,(K/k)\;\implies \sigma K=K$ .
3) I am guessing, and I am not sure I am right, that you mean $\;\sigma L/K\;$ being a solvable extension, which follows from the fact that $\;\sigma L=L\;$ as, again and as before, $\;L/K\;$ is normal. Now, if you meant $\;\sigma L/k\;$ then it could be it isn't even normal, so we don't have Galois and it is thus moot to talk about the automorphism group of the extension in this sense...if I understood correctly.