Is there a difference between the Galois group of $K/F$ and the Galois group of $E/K$ where $K$ is an intermediate field? Given an extension $E/F$, with intermediate fields $E/K_1/K_2/…../F$, I want to know if $Gal(K_n/F)=Gal(E/K_n)$. By Galois correspondence it seems like this should be true, there are still the same number and type of subgroups in the lattice but I wanted to make sure. 
So breaking down my question into parts:
1) Is this true ?
2) If it is only true sometimes what are the conditions ?
3)If it not true/sometimes true what is a counter example ?
4) If it is not true/sometimes true then how does one go about calculating $Gal(E/K_n)$ , and if it is true/sometimes true but it perhaps might be more practical to calculate it directly without considering Gal $(K_n/F)$, then how can one calculate it directly for example $Gal((\omega)/(\omega + \omega^{-1}))$ where $\omega$ is the 7th root of unity.?
 A: $\newcommand{\Gal}{\mathrm{Gal}}$This is not true. I'm going to drop your notation $K_1,\dots$ and just focus on a single intermediate field $E/K/F$, and I am going to assume the extension $E/F$ is Galois. 
Part of the usual statement of the Galois correspondence is that $(1)$ $E/K$ is always Galois, $(2)$ under the Galois correspondence $K$ corresponds to $\Gal(E/K)\le \Gal(E/F)$, and furthermore $(3)$ we have that $K/F$ is Galois $\iff$ $\Gal(E/K)$ is a normal subgroup of $\Gal(E/F)$, in which case we have $\Gal(K/F)\cong \Gal(E/F)/\Gal(E/K)$, with this isomorphism being established by the restriction $\Gal(E/F)\to\Gal(K/F)$ sending $\sigma\mapsto\sigma|_K$ (and this restriction makes sense precisely because $K/F$ is Galois) and noticing that the kernel is exactly $\Gal(E/K)$).
So you can see, asking for $\Gal(E/K)\cong\Gal(K/F)$ is basically the same as asking for a normal subgroup $N\triangleleft G=\Gal(E/F)$ such that $G/N\cong N$, which is not something you'd expect for most normal subgroups $N$ of a group $G$.
For a super trivial counter example, take your favorite nontrivial Galois extension $E/F$ and take $K=E$ so that $\Gal(E/K)=\Gal(E/E)$ is trivial but $\Gal(K/F)=\Gal(E/F)$ is not.
A: This is not true in general. Here’s a counter example.
Let $\xi$ be a primitive cube root of unity, and let $F =\mathbb{Q}(\sqrt[3]{2},\xi)$. Consider the Galis extension $F/\mathbb{Q}$ and note that its Galois group is $S_3$. 
Let $K$ be the fixed field of $A_3\le S_3$. Then the Galois group of $K/\mathbb{Q}$ is $A_3$, but the Galois group of $F/K$ is $\mathbb{Z}/ 2\mathbb{Z}$.
