Once you prove that the intersection of all subfields of a field is itself a field, you see that every field has a single smallest subfield. The function that Lubin suggests gives you a way to understand the structure of that smallest field. Lubin is defining the function $\varphi: \mathbb{Z} \rightarrow F$ (where $F$ is your original field) by $$\varphi(n) = \underbrace{1 + 1 + \dots + 1}_{\text{$n$ times}}.$$
Equivalently, you can also define this function as the group homomorphism $\varphi: (\mathbb{Z}, +) \rightarrow (F, +)$ with the property $\varphi(1) = 1_F$. (Since $1$ generates $\mathbb{Z}$ as a group, knowing where $\varphi$ sends $1$ is enough to define the whole function.)
At this point, you have to prove that $\varphi$ is also a ring homomorphism (which I'll leave to you).
Consider the kernel of $\varphi$. Since $\mathbb{Z}$ is a principal ideal domain, we have that $\ker \varphi = p\mathbb{Z}$ for some nonnegative integer $p$. This $p$ will, in fact, be the characteristic of the field $F$, meaning $p = 0$ or is a prime.
Now, suppose $p = 0$. Then consider the subset $$S = \left\{\varphi(a) \varphi(b)^{-1}: a, b \in \mathbb{Z}, b \not = 0\right\}$$ of $F$. You can show that $S$ is isomorphic to $\mathbb{Q}$. Since $\mathbb{Q}$ has no proper subfields, $S$ must be the smallest subfield of $F$.
On the other hand, suppose $p$ is a prime. Then $\varphi(\mathbb{Z}) \cong \mathbb{Z}/p\mathbb{Z}$. Since $\mathbb{Z}/p\mathbb{Z}$ has no proper subfields, $\varphi(\mathbb{Z})$ must be the smallest subfield of $F$.
Now, how do we know that a field $F$ with $p^n$ elements (where $p$ is prime) has characteristic $p$ (and hence a smallest subfield isomorphic to $\mathbb{Z}/p\mathbb{Z}$)? Since $F$ is a finite field, it must have some non-zero prime characteristic, $q$. Thus, $F$ has a subfield isomorphic to $\mathbb{Z}/q\mathbb{Z}$, and hence $F$ is a vector space over $\mathbb{Z}/q\mathbb{Z}$. Let $m = \dim F$ as a $\mathbb{Z}/q\mathbb{Z}$-vector space. Because $F$ is finite, $m$ must be finite as well. That means $p^n = q^m$. Hence $p = q$ and $m = n$.
(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
If $\alpha $ is a root of an irreducible polynomial $p(x)$ over some field $F $, then $F (\alpha)\cong \dfrac {F [x]}{(p (x))} $(a proof can be found in Dummit and Foote ).
For the first case,the roots of the polynomial $t^2-2$ are $\pm\sqrt2$. So $$\Bbb Q(\sqrt2)\cong\mathbb{Q}[t]/(t^2-2)\cong\Bbb Q(-\sqrt2). $$ But note that $\Bbb Q(\sqrt2)=\Bbb Q(-\sqrt2)$(why?). So there is only one field.
For $\mathbb{Q}[t]/(t^3-2)$, consider the roots of the polynomial $t^3-2$. The roots are $\sqrt [3]{2},\sqrt [3]{2}\omega, \sqrt [3]{2}\omega^2$, where $\omega$ is a non-trivial cube root of unity. By the same argument used above, we can see that $\Bbb Q(\sqrt [3]{2})\cong\Bbb Q(\sqrt [3]{2}\omega)\cong\Bbb Q(\sqrt [3]{2}\omega^2)$. Clearly $\Bbb Q(\sqrt [3]{2})\neq\Bbb Q(\sqrt [3]{2}\omega)$ and $\Bbb Q(\sqrt [3]{2})\neq\Bbb Q(\sqrt [3]{2}\omega^2)$. You can also show $\Bbb Q(\sqrt [3]{2}\omega)\neq\Bbb Q(\sqrt [3]{2}\omega^2)$ either by a proof by contradiction or by finding out the Galois group of $t^3-2$ over $\Bbb Q $.