Galois group of $(x^5-3)(x^5-7)$ over $\Bbb Q$ I am having some trouble with the following question:

Let $f(x)=(x^5-3)(x^5-7)\in\Bbb Q[x]$. Find the degree of the splitting field of $f$ over $\Bbb Q$, and determine the Galois group of $f$ over $\Bbb Q$.

I noticed that $x^5-3$ and $x^5-7$ are irreducible in $\Bbb Q[x]$, and so their Galois groups are isomorphic to a transitive subgroup of $S_5$. I also found that the splitting fields of these two quintics are $\Bbb Q(\sqrt[5] 3,\alpha)$ and $\Bbb Q(\sqrt[5] 7,\alpha)$ where $\alpha^5=1$. Since $\alpha$ has degree $4$ over $\Bbb Q$ and $\sqrt[5] 3,\sqrt[5] 7$ have degree $5$ over $\Bbb Q$, we have that the degrees of these splitting fields are $20$. Hence, we have $$\operatorname{Gal}(x^5-3/\Bbb Q)\cong F_{20}\cong \operatorname{Gal}(x^5-7/\Bbb Q) $$
But now I'm supposed to use a theorem we proved that $\operatorname{Gal}(f/\Bbb Q)$ is a subgroup of the direct product of the above Galois groups. I am not sure how to do this: i.e., $\color{red}{\text{how to determine which subgroup $\operatorname{Gal}(f/\Bbb Q)$ corresponds to}}$. 
For the degree of the splitting field of $f$ over $\Bbb Q$, I argued that it is $100$ since the compositum $\Bbb Q(\sqrt[5] 7,\alpha)\Bbb Q(\sqrt[5] 3,\alpha)=\Bbb Q(\sqrt[5] 3,\sqrt[5] 7,\alpha)$ has degree $5\cdot 5\cdot 4=100$ over $\Bbb Q$. $\color{red}{\text{Does this make sense}}$?

I also had another question to determine the Galois group of $(x^3-2)(x^4-2)$ over $\Bbb Q$. I showed that $x^3-2$ has Galois group $S_3$ and $x^4-2$ has Galois group $D_4$. Then I argued that the intersection of the two splitting fields, $\Bbb Q(\sqrt[3]2,\omega)\cap\Bbb Q(\sqrt[4]2,i)$ is just $\Bbb Q$, so we have $\operatorname{Gal}((x^3-2)(x^4-2)/\Bbb Q)\cong S_3\times D_4$. $\color{red}{\text{Does this make sense}}$? In order to show $\Bbb Q(\sqrt[3]2,\omega)\cap\Bbb Q(\sqrt[4]2,i)=\Bbb Q$ I just argued that $\sqrt[4]2,i\notin\Bbb Q(\sqrt[3]2,\omega)$, $\color{red}{\text{is this sufficient}}$?
 A: On the first "Does this make sense?": Yes, it pretty much does. Using the tower law, we get that $$[\mathbb{Q}(\sqrt[5]3,\sqrt[5]7,\alpha):\mathbb{Q}]=[\mathbb{Q}(\sqrt[5]3,\sqrt[5]7,\alpha):\mathbb{Q}(\sqrt[5]7,\alpha)][\mathbb{Q}(\sqrt[5]7,\alpha):\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}]=5*5*4=100.$$
On your second "Does this make sense?": Yes, it also does. If $K_1$ and $K_2$ are Galois extensions and $K_1 \cap K_2=F$, $$Gal(K_1K_2/F)\cong Gal(K_1/F)\times Gal(K_2/F).$$ So, if $K_1=\mathbb{Q}(\sqrt[3]2,\omega)$ and $K_2=\mathbb{Q}(\sqrt[4]2,i)$ and $Gal(K_1/\mathbb{Q})\cong S_3$ and $Gal(K_2/\mathbb{Q})\cong D_4$, we get the result you got. However, I think that you could be more complete when arguing that $K_1\cap K_2=\mathbb{Q}$, explaining it a bit more- even though you're not wrong.
On the first part, as we argued in the comment section, $Gal(\mathbb{Q}(\sqrt[5]3,\sqrt[5]7,\alpha)/\mathbb{Q})\cong C_5^2\times C_4$. This is due to the fact that, once $\mathbb{Q}(\sqrt[5]3,\sqrt[5]7)\cap \mathbb{Q}(\alpha)=\mathbb{Q},$ 
$$Gal(\mathbb{Q}(\sqrt[5]3,\sqrt[5]7,\alpha)/\mathbb{Q})\cong Gal(\mathbb{Q}(\sqrt[5]3,\sqrt[5]7)/\mathbb{Q})\times Gal(\mathbb{Q}(\alpha)/\mathbb{Q})\cong Gal(\mathbb{Q}(\sqrt[5]7)/\mathbb{Q})\times Gal(\mathbb{Q}(\sqrt[5]3)/\mathbb{Q})\times Gal(\mathbb{Q}(\alpha)/\mathbb{Q})\cong C_5\times C_5 \times C_4=C_5^2\times C_4 $$
