Computing the degree of a Galois extension. 
Let $K = \Bbb{Q}(3^{1/5}, 3^{1/3})$. Compute $[K: \Bbb{Q}]$ (degree of the extention). Find $\alpha$ (not unique) such that $F=\Bbb{Q}(3^{1/5}, 3^{1/3} , \alpha)$ is the smallest Galois extension containing $\alpha$ and $K$. Finally, compute $[F:\Bbb{Q}]$ and the order of $Gal_{\Bbb{Q}}(F)$.

We know that $$[K:\Bbb{Q}]=[K: \Bbb{Q}(3^{1/5})][\Bbb{Q}(3^{1/5}): \Bbb{Q}].$$
$x^5-3$ is irreducible over $\Bbb{Q}$ by Eisenstein's criteria, so $[\Bbb{Q}(3^{1/5}): \Bbb{Q}]=5$.
Since $3^{1/3} \not\in \Bbb{Q}(3^{1/5})$, $[K: \Bbb{Q}(3^{1/5})]$ is at least 2. Since $3^{1/5}$ is a root of $x^3-3$ over $\Bbb{Q}(3^{1/5})$, the degree is at most 3. But the roots of $x^3-3$ are $3^{1/3}$, $\zeta_33^{1/3}$, and $\zeta^2_33^{1/3}$. If $x^3-3$ was reducible, it would have a linear term so one of those roots would be in $\Bbb{Q}(3^{1/5})$, a contradiction. So $[K: \Bbb{Q}(3^{1/5})]=3$, and $[K:\Bbb{Q}]=15$. 
I'm not sure if I clearly understand the second question, but if there are no restrictions on $\alpha$, then can't we just choose $\alpha$ to be in $K$? Because that would obviously be the smallest Galois extension over K, right? And $[F:\Bbb{Q}]=Gal_{\Bbb{Q}}(F) = 15$. 
I feel like there is something wrong with my answer, because...
Since $[K:\Bbb{Q}]=15$, the minimal polynomial of $3^{1/3}$ and $3^{1/5}$ must be of degree 15. But the polynomial $(x^5-3)(x^3-3)$ also has these roots and is of degree less than 15. So I think there is something wrong with my answer. 

EDIT

In order to make it a Galois extension, we need to add $\zeta_3$ and $\zeta_5$ to $K$. So $K(\zeta_3, \zeta_5)$ is the smallest extension of $K$ that will make it Galois. 
By the theorem of the primitive element, we know that there is an $\alpha$ such that $K(\alpha) = K(\zeta_3,\zeta_5)$. 
Let $\alpha=\zeta_3\zeta_5$. Since $\zeta_3\zeta_5 \in K(\zeta_3,\zeta_5)$, and since $(\zeta_3\zeta_5)^6 = \zeta_5 \in K(\alpha) \implies \zeta_3 \in K(\alpha)$; we know that $K(\alpha) = K(\zeta_3, \zeta_5)$. 
Now we need to compute $[F:K]$. We already know the degree of $K$ over $\Bbb{Q}$, so we just need to compute $[K(\alpha):K]$ or $[K(\zeta_3, \zeta_5):K]$.
First we compute $[K(\zeta_3):K]$. We know that $\zeta_3$ is a root of $x^3-1$ over $K$. However, $x^3-1$ is reducible, since $-1$ is in $K$. We know that $x^3-1 = (x-1)(x^2+x+1)$. So $\zeta_2$ is also a root of $x^2+x+1$. Now, if $x^2+x+1$ wasn't the minimal polynomial of $\zeta_3$, then it would be reducible to linear factors and $\zeta_3$ would be in $K$, a contradiction. So $[K(\zeta_3):K]=2$. 
We now have to compute $[K(\zeta_3, \zeta_5):K(\zeta_3)]$. The minimal polynomial must divide $x^5-1$. Since $-1$ is in $K(\zeta_3)$, we have $(x^5-1)=(x-1)(x^4+x^3+x^2+x+1)$. We cannot reduce $x^4+x^3+x^2+x+1$ into polynomials of degree 3 or 1, since then we would need to have $\zeta_5 \in K(\zeta_3)$. But now I have to check that we can't reduce it into polynomials of degree two, and I'm stuck...
Thank you in advance 
 A: You have made significant progress. 
I would approach the remaining part (may be also the earlier parts) using the following, often useful observation. If you have several intermediate fields $\mathbb{Q}\subset E_i\subset F$, $i=1,2,\ldots,m$, such that the extension degrees are $n_i=[E_i:\mathbb{Q}]$, then 
$$
[F:\mathbb{Q}]\ge l.c.m.(n_1,n_2,\ldots,n_m).
$$
This follows from the multiplicativity of extension degrees in a tower of fields, because that implies that the extension degree $[E:\mathbb{Q}]$ is a multiple of all the $n_i$:s, hence a multiple of their least common multiple.
This is particularly useful, when the integers $n_i$ are pairwise coprime, as then their least common multiple equals their product.
Here you have showed that with $E_1=\mathbb{Q}(3^{1/3})$ you have 
$n_1=[E_1:\mathbb{Q}]=3$, and that with $E_2=\mathbb{Q}(3^{1/5})$ you have
$n_2=[E_2:\mathbb{Q}]=5$. Now I'm asking you to concentrate on proving that
with $E_3=\mathbb{Q}(\zeta_3,\zeta_5)=\mathbb{Q}(\alpha)$ you get
$n_3=[E_3:\mathbb{Q}]=8$. As $3,5$ and $8$ are coprime, this gives you something useful :-)
There are several ways to prove $n_3=8$. The simplest route depends on what bits you have covered. I will describe three ways of doing it. They are ordered by decreasing need for background information, but unfortunately also by an increasing need for trickstery.
If you have covered the general result that
$$
[\mathbb{Q}(\zeta_n):\mathbb{Q}]=\phi(n),
$$
the value of the Euler totient function, then you can use that by proving that
$\alpha=\zeta_{15}^8$ and $\zeta_{15}$ generate the same extension of $\mathbb{Q}$. Note that there is no need to do this over the more complicated field $K$ - that's why the above lower bound is so useful in this case!
Another approach (hinted at by Ted Shifrin) relies on some bits of Galois theory. If you know that the Galois group $G$ of the extension $\mathbb{Q}(\zeta_5)/\mathbb{Q}$ is cyclic of order four, with the generating automorphism $\sigma$ defined by $\sigma(\zeta_5)=\zeta_5^2$, then we can use that to show
that $\zeta_3\notin\mathbb{Q}(\zeta_5)$. You have already shown that 
$[\mathbb{Q}(\zeta_3):\mathbb{Q}]=2$. So if we had $\zeta_3\in\mathbb{Q}(\zeta_5)$, then Galois theory tells us that $\zeta_3$ would have to be fixed by $\sigma^2$, because $\sigma^2$ generates the only subgroup of index two of $G$.
But you can prove that $\sigma^2$ is the restriction of the usual complex conjugation to $\mathbb{Q}(\zeta_5)$. Leaving that bit to you for the time being at least, as you are manifestly not scared of hard work!! So the numbers fixed under $\sigma^2$ are all real, hence $\zeta_3$ is not one of them. With that bit of information you can then show that
$[\mathbb{Q}(\zeta_5,\zeta_3):\mathbb{Q}(\zeta_5)]=2$, and deduce that
$n_3=8$.
And finally, if you haven't covered the necessary bits to the two approaches above, then there is always, the oft neglected, brute force approach. You know that the minimal polynomial of $\zeta_5$ over $\mathbb{Q}(\zeta_3)$ must be a factor of
$$
\phi_5(x)=x^4+x^3+x^2+x+1=(x-\zeta_5)(x-\zeta_5^2)(x-\zeta_5^3)(x-\zeta_5^4).
$$
You already ruled out the possibility of a linear factor, and the remaining task was to check that no quadratic factor of $\phi_5(x)$ has coefficients in $\mathbb{Q}(\zeta_3)$. It suffices to check the polynomials $(x-\zeta_5)(x-\zeta_5^j)$ with $j=2,3,4$. Here we observe that
$(x-\zeta_5)(x-\zeta_5^4)$ has real irrational coefficients, so that won't do.
The other two candidate factors have a constant term that is a non-trivial power of $\zeta_5$. As you have already showed that $\zeta_5\notin\mathbb{Q}(\zeta_3)$,
this will do it. Again leaving the details to you.
With knowledge of all $n_1,n_2,n_3$ you get a lower bound for $[F:\mathbb{Q}]$. You should also be able to prove that the lower bound is equal to the easy upper bound. That will settle the question.
