Subfields of $\mathbb{Q}(3^{1/16})$ 
The degree of $\mathbb{Q}(\alpha^8)$ over $\mathbb{Q}$ would be 2, because the minimum polynomial of $\alpha^8$ is $x^2-3$ which has degree 2.  To compute $[\mathbb{Q}(\alpha^4):\mathbb{Q}(\alpha^8)]$ we first compute $[\mathbb{Q}(\alpha^4):\mathbb{Q}]$.  This is 4 because the minimal polynomial is $x^4-3$. So $[\mathbb{Q}(\alpha^4):\mathbb{Q}(\alpha^8)]$ is 2, by tower law.  Similarly, we get that the degrees of the other intermediate extensions are 2.  To check that these types of polynomials are irreducible, we just use eisenstein's criterion with $p = 3$.
Now for the second part of the question I'm not completely sure. An intermediate field of $F$ must be generated by some $\alpha^n$ since if it doesn't contain any, then it can only be $\mathbb{Q}$.  If $n$ does not divide 16, then the minimal polynomial has degree greater than 16(I think), which means it can't be a subfield.  But how does this argument require using the constant term?
Source: https://dornsife.usc.edu/assets/sites/363/docs/F17_510ab.pdf 
 A: As mentioned in the comments the subfields aren't neccesarily of the form $\mathbb{Q}(\alpha^{\frac 1n})$. Nevertheless the easiest way would be to consider the Galois closure of the field and use Galois correspondence.
It's not hard to compute that $\mathbb{Q}(\alpha,\zeta_{16})$, where $\zeta_{16}$ is the principal 16-th root of unity, is the Galois closure of $\mathbb{Q}(\alpha)$. Also we have that $x^{16}-3$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}(\zeta_{16})$. This follows as $3$ is a prime in the ring of integers $\mathbb{Z}[\zeta_{16}]$ and thus we can use the generalized Eisenstein Criterion. Now it's not hard to see that the Galois group is generated by:
$$\tau:
     \begin{array}{lr}
       \alpha \to \zeta_{16}\alpha\\
       \zeta_{16} \to \zeta_{16}
     \end{array}
\quad \quad   \sigma:
     \begin{array}{lr}
       \alpha\to \alpha\\
       \zeta_{16} \to \zeta_{16}^3
     \end{array}
\quad \quad   \pi:
     \begin{array}{lr}
       \alpha\to \alpha\\
       \zeta_{16} \to \zeta_{16}^7
     \end{array}$$
In fact we have that the Galois group is isomorphic to the semidirect product $\mathbb{Z}_{16}\rtimes \left(\mathbb{Z}_2 \times \mathbb{Z}_4\right)$. As $\{e\} \rtimes \left(\mathbb{Z}_2 \times \mathbb{Z}_4\right)$ is the subgroup fixing $\mathbb{Q}(\alpha)$ you need to find all subgroups of $\mathbb{Z}_{16}\rtimes \left(\mathbb{Z}_2 \times \mathbb{Z}_4\right)$ containing it. They are exactly:
$$\langle e \rangle \rtimes \left(\mathbb{Z}_2 \times \mathbb{Z}_4\right), \langle 8 \rangle \rtimes \left(\mathbb{Z}_2 \times \mathbb{Z}_4\right),\langle 4 \rangle \rtimes \left(\mathbb{Z}_2 \times \mathbb{Z}_4\right),\langle 2 \rangle \rtimes \left(\mathbb{Z}_2 \times \mathbb{Z}_4\right),\langle 1 \rangle \rtimes \left(\mathbb{Z}_2 \times \mathbb{Z}_4\right)$$
Thus by Galois correspondence $\mathbb{Q} \subset \mathbb{Q}(\alpha)$ has exactly $5$ intermediate subfields and they must be the one already mentioned.
Hence the proof.
A: I will add a solution following the hint without using Galois theory.
Hint: The key idea is that given any intermeidate field $L$, if 
$$\big|L:\mathbb{Q}\big| = \big|\mathbb Q[\alpha^n] : \mathbb{Q}\big|,$$ 
to show $L = \mathbb{Q}[\alpha^n]$, we only need to show $\alpha^n \in L$, otherwise if $L\supsetneq \mathbb{Q}[\alpha^n]$, we get a contradiction from tower law and their degrees.
And we will get $\alpha^n \in L$ by looking at the constant term in the minimal polynomial of $\alpha$ in $L[x]$. 
Solution: Denote the polynomial
$$p(x) = x^{16}-3 = (x-\alpha)(x-\zeta\alpha) \cdots (x-\zeta^{15}\alpha)$$
where $\zeta$ is the 16th root of unity and $\alpha =  3^{1/16}$. 
Now Given a intermeidate field $L$ and $|L:\mathbb{Q}| = 16/n$, $n$ could be $2,4,8$. By tower law, we have 
$$\big|\mathbb{Q}[\alpha]:L\big| = \big|\mathbb{Q}[\alpha]:\mathbb{Q} [\alpha^n] \big| = n$$
Look at the minimal polynomial of $\alpha$ in $L[x]$, call this $g(x)$. Observe that $p(\alpha) = 0$, so we must have $g | p$ in $L[x]$. So this degree $n$ polynomial $g(x)$ can be factored as 
$$g(x) = (x-\zeta^{k_1}\alpha)\cdots(x-\zeta^{k_n}\alpha) $$
 and $\{k_1, \cdots, k_n\}$ is just some subset of $\{0,1,2,\cdots, 15\}$.
Note the constant term of $g(x)$ is $\zeta^{k_1+k_2+\cdots k_n} \alpha^nm \in L$. Since $L $ is a real field, we must have $\zeta^{k_1+k_2+\cdots k_n} = \pm1$, so $\alpha^n \in L$ and we are done.
