$KE / E$ is Galois: Use Proposition 17012 and (3) of Proposition 2205 and the premise that $K / F$ is Galois.

$\mathrm{Gal}(KE / E) \cong \mathrm{Gal}(K / K \cap E)$: The map $$ \mathrm{Gal}(KE / E) \to \mathrm{Gal}(K / K \cap E), \qquad \sigma \mapsto \sigma |_{K} $$ is a group homomorphism. The kernel is $\{ \operatorname{id}_{KE} \}$. Let $H$ be its image. Then $$ \begin{aligned} \alpha \in K^{H} &\iff \alpha \in K \text{ and } \sigma(\alpha) = \alpha \text{ for all } \sigma \in \mathrm{Gal}(KE / E) \\ &\iff \alpha \in K \cap \left( KE \right)^{\mathrm{Gal}(KE / E)}. \end{aligned} $$ $KE / K$ is Galois $\implies (KE)^{\mathrm{Gal}(KE / E)} = E$. Hence, $$ \alpha \in K^{H} \iff \alpha \in K \cap E. $$ It follows that $K^{H} = K \cap E$. Then again by the Fundamental Theorem of Galois Theory, we have $$ H = \mathrm{Gal}(K / K^{H}) = \mathrm{Gal}(K / K \cap E). $$