(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.
I agree with your choice of splitting field, although I think your solution is probably a bit longer than is standard. Just note that, if $E$ is the splitting field of $f$ over $\mathbb{Q}$, then $\sqrt{2}, \sqrt{3}, \sqrt{5} \in E$, so $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}) \subseteq E$. In fact $f$ splits in $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$, which means that $E = \mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$.
Now, $E$ is a splitting field over $\mathbb{Q}$, which means that (by a general result of Galois Theory) it is a normal extension of $\mathbb{Q}$. Furthermore, $\mathbb{Q}$ is a field of characteristic zero, which means that any finite field extension of $\mathbb{Q}$ is separable (fields of characteristic zero are perfect), so $E/\mathbb Q$ is a normal, separable extension, hence it is a Galois extension.
Since $E/\mathbb{Q}$ is a finite Galois extension, its Galois group $G = \text{Gal}(E/\mathbb{Q})$ has $\lvert G \rvert = [E:\mathbb Q]$, and by the linked question, $[E:\mathbb{Q}] = 8$, so $G$ is a group of order $8$. This means that there are eight $\mathbb{Q}$-automorphisms of $E$. Let $\varphi:E\to E$ be such a $\mathbb{Q}$-automorphism. For each of $d = 2,3,5$, we have $\varphi(\sqrt{d})^2 = \varphi(d) = d$, which means that $\varphi(\sqrt{d}) = \pm \sqrt{d}$. There are thus $2^3 = 8$ choices for the images of the $\sqrt{d}$ under $\varphi$. Note that each such choice determines $\varphi$ fully because $E$ is generated over $\mathbb{Q}$ by these $\sqrt{d}$. Since there are eight such choices, and we know that $\lvert G \rvert = 8$, so in fact the elements of $G$ are precisely the $\mathbb{Q}$-automorphisms induced by
$$
(\sqrt{2},\sqrt{3},\sqrt{5}) \mapsto (\epsilon_1\sqrt{2},\epsilon_2\sqrt{3},\epsilon_3\sqrt{5})
$$
for all possible choices of $\epsilon_i =\pm 1$.
If we identify the roots of $f$ with the set $\Omega = \{1,2,3,4,5,6\}$ via
$$
(\sqrt{2}, -\sqrt{2},\sqrt{3},-\sqrt{3},\sqrt{5},-\sqrt{5}) \leftrightarrow (1,2,3,4,5,6)
$$
then $G$ may be viewed as a subgroup of the symmetric group $S_6$ generated by the transpositions $(12),(34),(56)$, which is isomorphic to $C_2\times C_2\times C_2$.
I have tried to present the argument as rigorously as possible, in keeping with the thorough nature of your solution to the first part. However, I think it's also important to note that the group can be seen in a much simpler, more intuitive way. Basically, $\sqrt{2},\sqrt{3},\sqrt{5}$ are generators of $E$, and they are in some sense independent. Also each one must be mapped by an element of $G$ to either itself, or minus itself (its additive inverse). As such, pretty much all an element of the Galois group does is "flips" some of these square roots. We have three objects to flip (think three levers that can be toggled between two states), so the group is $C_2\times C_2 \times C_2$.
Best Answer
The Galois group is actually $S_3$. As you correctly note, the fact that a cubic and a quadratic which are irreducible over $\mathbb Q$ split over $E$ implies that $E/\mathbb Q$ has subextensions of degree $2$ and $3$, so must have degree at least $\mathrm{lcm}(2,3)=6$. Thus $|\mathrm{Gal}(E/\mathbb Q)|\ge 6$. On the other hand, since the Galois group of the splitting field of $f(x)$ acts faithfully on the roots of $f$ (i.e. it permutes the roots, and the only element which fixes every root is the trivial automorphism), we have that $\mathrm{Gal}(E/\mathbb Q)$ embeds in $S_3$. But since $|S_3|=6$, $\mathrm{Gal}(E/\mathbb Q)$ must be the whole group so $\mathrm{Gal}(E/\mathbb Q)\cong S_3$.