So I was doing a past exam and am kind of stuck on this question, I know how to do part a and b but unsure on how to proceed with the remaining ones. Any help on those will be appreciated along with any hints. Thanks
EDIT
Here are my workings for the (a) and (b)
(a)
okay, for (a) we look at the roots for $x^7$ - 3, which are $\sqrt[7] {3}, \zeta_7 \sqrt[7]{3}, \zeta_7^2 \sqrt[7]{3}, \ldots , \zeta_7^6 \sqrt[7]{3}$ where $ \zeta_7 = e^((2i\pi)/7)$. now we claim that K = $Q(\zeta_7, \sqrt[7] {3})$, we show $Q(\zeta_7, \sqrt[7] {3}) \subset K$ and vice versa as $\sqrt[7] {3} \in K$ and $\zeta_7 = \zeta_7 (\sqrt[7] {3}/\sqrt[7] {3})$ and the other way is true because all of the roots are contained in $Q(\zeta_7, \sqrt[7] {3})$
As for the degrees, for $ \sqrt[7] {3}$ is 7 and for $\zeta_7$ is 6 and thus [K : Q] = 42
for (b), we know K/Q has finite elements hence it is Galois
For (b), "has finite elements" isn't really what Galois means. For instance, $\Bbb Q(\sqrt[7]3)$ is not Galois. It is Galois because it is separable (this comes automatically in characteristic $0$) and normal (it is the splitting field of a polynomial, in this case of $x^7-3$).
For (c), the third Sylow theorem lets you conclude quite easily how many subgroups there are of order $7$: There is one. The corresponding intermediate field would be one which has order $6$ over $\Bbb Q$. If your answer to (a) was $\alpha = \sqrt[3]7$ and $\beta = e^{2\pi i/7}$, then it should be rather easy to guess what this field is.
For (d), you have the following two maps: $$ \sigma: \alpha\mapsto \alpha\beta\quad \beta\mapsto\beta \\ \tau:\alpha\mapsto \alpha\quad\beta\mapsto \beta^3 $$ They are rather standard when it comes to the Galois group of $\Bbb Q(\sqrt[n]{k}, e^{2\pi i/n})$. The exponent $^3$ in $\tau$ must be chosen with some care so that $\tau$ doesn't get lower order than $n-1$ (for instance, $\beta\mapsto \beta^2$ has order $3$).
For (e), that's just straight-up calculation (note that $\tau^{-1}(\beta) = \beta^{5}$): $$ \tau\sigma\tau^{-1}(\alpha) = \tau\sigma(\alpha) = \tau(\alpha\beta) = \alpha\beta^3= \sigma^3(\alpha)\\ \tau\sigma\tau^{-1}(\beta) = \tau\sigma(\beta^5) = \tau(\beta^5) = \beta $$ where that last one also fits with $\sigma^3(\beta)$. So we have $\tau\sigma\tau^{-1} = \sigma^3$.
Finally, for (f), let $N = \langle \sigma\rangle$ and $H = \langle \tau\rangle$. Then our above result for (e) gives $G = N\rtimes_\varphi H$, with $\varphi:H\to\operatorname{Aut}(N)$ given by $\varphi(\tau) = \square\,^3$, the cubing automorphism.