There's a general tool to deal with this type of extensions. Suppose you have a diamond of extensions $F \subseteq L,L' \subseteq E$ (make the drawing) where $L\cap L'=F,LL'=E$.
Suppose moreover that $L/F$ is normal, and $E/F$ is Galois. Let $G={\rm Gal}(E/F)$, $H={\rm Gal}(E/L)$ and $K={\rm Gal}(E/L')$. Then show that $HK=G, H\cap K=1$ and $K\lhd G$. All this means that $G$ is the semidirect product of $H$ and $K$, in particular every element $\sigma$ of $G$ is written uniquely as $\tau\rho$ with $\tau\in H,\rho\in K$. In your case one gets $G$ is a semidirect product $C_2\rtimes C_4$ which is the dihedral group of order four. But one gets more than this, the above gives an explicit "decomposition" of the Galois group in terms of the subextensions $L,L'$.
To be more precise, take $L=\Bbb Q(i),L'=\Bbb Q(7^{1/4}),F=\Bbb Q$ and $E=LL'$ (your extension). Then $L/\Bbb Q$ is normal and $L\cap L'=\Bbb Q$ (this is not hard to prove, since $L$ is obtained by adjoining $i$, and its a degree two extension). One can see that $E/L$ is of degree four and $E/L'$ is of degree two.
The technique above is very useful when one adjoins more than two elements, since one can iterate the decomposition. For example, one can explicitly calculate the Galois group of $(X^4-3)(X^3-5)$ over $\Bbb Q$ as a semidirect product $(C_3\rtimes C_4)\rtimes C_2$ explicitly: it is generated by $\omega,\eta,\psi$ such that $\omega^3=\eta^4=\psi^2=1$
- $\eta(3^{1/4})=3^{1/4}i$ and $\eta$ fixes $i,5^{1/3}$
- $\psi(i)=-i$ and $\psi$ fixes $3^{1/4},5^{1/4}$,
- $\omega(5^{1/3})=5^{1/3}\xi$ and $\omega$ fixes $i,3^{1/4}$.
Here $\xi$ a primitive cuberoot of $1$, and every element of $G$ is written uniquely as $\omega^i\eta^j\psi^k$ with $i=0,1,2,j=0,1,2,3,k=0,1$. Moreover $\psi\omega\psi =\omega^{-1},\eta\omega\eta^{-1}=\omega^{-1}$. A bit more calculations can give you a presentation for $G$, which has order $24$.