Let $ H_i $ denote the isomorphic copy of $ S_{n-1} $ in $ S_n $ composed of elements which fix $ i $. For $ L/K $ a Galois extension with Galois group $ S_n $, consider the fixed field $ F_1 $ of $ H_1 $. Clearly $ [F_1 : K] = n $, and $ F_1/K $ is separable since it is a subextension of the separable extension $ L/K $. Pick a primitive element $ \alpha_1 $ for this extension. Then, $ \alpha_1 $ has exactly $ n $ $ K $-conjugates (counting itself), say $ \alpha_1, \alpha_2, \ldots, \alpha_n $, since its stabilizer in the Galois group is exactly $ H_1 $, which is a subgroup of index $ n $ in the Galois group. The subgroup of $ S_n $ fixing $ K(\alpha_1, \alpha_2, \ldots, \alpha_n)/K $ is exactly
$$ \bigcap_{1 \leq i \leq n} H_i = \{ \textrm{id} \} $$
By Galois correspondence, it follows that $ K(\alpha_1, \alpha_2, \ldots, \alpha_n) = L $. Therefore, $ L $ is the splitting field of the minimal polynomial of $ \alpha_1 $ over $ K $, which is a polynomial of degree $ n $.
(b) take products and sums of the generators in $K$ to get elements in $E$. E.g $\sqrt{1 + \sqrt{2}}$ is contained in $K$, so $1 + \sqrt{2}$ is. Similarly $1 - \sqrt{2}$ is. Subtracting these two elements $1$ and $\sqrt{2}$ are both contained in $K$.
(c) When you say $E \sqrt{1 + \sqrt{2}}$ you must mean $L = E ( \sqrt{ 1 + \sqrt{2}})$. Since $E \subset K$, we can use degrees here to show they are the same (if a finite field extension is contained in another finite field extension of the same degree then they are the same. $L/E$ has degree $2$ and $E/\mathbb{Q}$ has degree $4$, as $E / \mathbb{Q}(\sqrt{2})$ has degree $2$ and $\mathbb{Q}(\sqrt{2}) / \mathbb{Q}$ has degree $2$. Now show that $K$ has degree $8$ using the tower $K / \mathbb{Q}(\sqrt{1 + \sqrt{2}})/\mathbb{Q}(\sqrt{2}) / \mathbb{Q}$.
(d) For a simple extension $L(a)/L$ of fields, with $a$ having minimal polynomial $f(x) \in L[x]$, there is always an automorphism permuting any two roots of $f$ contained in $L[x]$. To get one of the desired maps in the problem, take the map $\mathbb{Q}(\sqrt{2}) \rightarrow \mathbb{Q}(\sqrt{2})$ sending the root $\sqrt{2}$ of the minimal polynomial $x^2 - 2$ to the other root $-\sqrt{2}$ of the minimal polynoimal. This gives an automorphism of $\mathbb{Q}(\sqrt{2})$. This gives a map $\mathbb{Q}(\sqrt{2}) \rightarrow \mathbb{Q}(\sqrt{2}, i)$, which extends to a map $\mathbb{Q}(\sqrt{2}, i)$ sending $i$ to $i$. The reason this extension exists is that $\mathbb{Q}(\sqrt{2}, i) / \mathbb{Q}(\sqrt{2})$ is a simple extension generated by $i$, with minimal polynomial $x^2 + 1$, and $x^2 + 1$ has a root in the target of the embedding $\mathbb{Q}(\sqrt{2}) \rightarrow \mathbb{Q}(\sqrt{2}, i)$.
So the two general facts we used were
Let $L(a)/L$ be a finite simple extension of fields. For any root $b$ of the minimal polynomial $f$ in $L(a) / L$, there is an $L$-automorphism of $L(a)$ sending $a$ to $b$.
Let $L(a)$ be a finite simple extension of fields. If $M/L$ contains a root $b$ of the minimal polynoimal $f(x)$ of $a$ in $L[x]$, then there is an embedding $L(a) \rightarrow M$ sending $a$ to $b$.
(e) this uses the same methods discussed in the last one. Here $K/E$ is a simple extension with $a = \sqrt{1 + \sqrt{2}}$ its generator. To proceed, calculate the minimal polynomial $f(x)$ of $a$ over $E$. There is an $E$-automorphism of $K$ sending $a$ to the other root of $f(x)$ in $K$.
(f) you got it!
(g) $fg$ and $gf^3$ are determined by what they do to the roots of certain polynomials. This is nice for us, since we have identified finitely many of these, and automorphisms send roots to other roots. So it is simply a matter of evaluating what they do to the roots. E.g. $f g ( \sqrt{1 + \sqrt{2} } ) = f ( \sqrt{1 + \sqrt{2}} ) = \sqrt{1 - \sqrt{2}}$. Do all the calculations of this kind, observe that they are the same for $fg$ and $gf^3$, and you'll be done.
When we found roots of the minimal polynomials and observed how each automorphism permutes them, we have really reduced an "intractible" question of automorphisms of fields to a tractible question of a finite group acting on the roots; automorphisms are determined by how they act on the roots of certain polynomials, and they permute those roots. In this way, instead of automorphisms acting on a field we have automorphisms acting on a finite set. The second should seem a lot easier to answer, since it is about something finite- in the worst case, we have a brute force approach.
(h) This is potentially the most tricky part, but it's also where all this computation has been leading us. $D_8$ is given by generators and relations as $\langle x, y | x^4, y^2, xy = yx^3 \rangle$. To give an isomorphism, let's start with a map from the free group $\langle x, y \rangle$ to the galois group in question sending $x$ to $f$ and $y$ to $g$. We must show a few things to establish an isomorphism:
$x^4$ is sent to $\text{Id}$, and so are $y^2$ and $x^{-3}y^{-1}xy$. You showed the last one already, but the first two are not so different!
All automorphisms in the galois group are some combination of $f$ and $g$.
The order of the galois group is $8$.
I have to go for now, but I hope this is enough.
Best Answer
To clarify, let's think of an example. If $f(x)=x^2-2,$ then $K=\mathbb{Q}(\sqrt{2})$ and $f(x^2)=x^4-2$. This polynomial decomposes over $K$ as $(x^2-\sqrt{2})(x^2+\sqrt{2})$. $L$ is obtained as a successive quadratic extension $$K\subset K(\sqrt[4]{2})\subset K( \sqrt[4]{2},i\sqrt[4]{2})=K(\sqrt[4]{2},i)=L$$ where $i=\sqrt{-1}$. We see that $$\mathrm{Gal}(L/K)=(\mathbb{Z}/2\mathbb{Z})^2.$$
The general case is the same, since $f(x^2)$ decomposes in $K$ as $$(x^2-a_1)(x^2-a_2)\cdots(x^2-a_d)$$ and $L$ is given by $$L=K(\sqrt{a_1},\sqrt{a_2},\cdots\sqrt{a_d})$$ where some of the square roots can be redundant. The Galois group $\mathrm{Gal}(L/K)$ is the set of involutions which negates appropriate $\sqrt{a_i}$'s (*), so it is in fact an abelian elementary 2-group $(\mathbb{Z}/2\mathbb{Z})^e$ as stated.
Note: An example of redundant square root, thus the case $e<d$ arises for $f(x)=x^2+1$.
(*)EDIT (explanation): Let us throw those $\sqrt{a_i}$ which is already in $K(\sqrt{a_1},\dots,\sqrt{a_{i-1}})$ away and remove redundancy. By renumbering, we just denote the resulting generators of $L$ over $K$ as $$L=K(\sqrt{a_1},\sqrt{a_2},\cdots\sqrt{a_e}).$$ Thus, we can assume that the successive quadratic extensions $$K\subset K(\sqrt{a_1})\subset K(\sqrt{a_1},\sqrt{a_2})\subset \cdots\subset K(\sqrt{a_1},\dots,\sqrt{a_e})=L$$ are all strict inclusions. In particular, we see $[L\colon K]=2^e$ and $L/K$ is Galois by construction.
Now, for every $\sigma\in\mathrm{Gal}(L/K)$ and every $i=1,\dots,e$ we have $\sigma(\sqrt{a_i})=\pm\sqrt{a_i}$ since they are the only solutions of the minimal polynomial $x^2-a_i$ over $K$. In particular, we see that $\sigma^2=\mathrm{id}_L$ for every $\sigma$. This is enough to conclude that $\mathrm{Gal}(L/K)$ is elementary abelian of order $2^e$. Perhaps more directly, it is also clear that we have $\sigma\sigma'=\sigma'\sigma$ for $\sigma,\sigma'\in \mathrm{Gal}(L/K)$. Therefore, we get the conclusion again. (Note that, a counting of elements shows that the choices of $\pm$ on $i$ can be done free of conditions, to make up $2^e$ elements.)