extension field - Galois group of the $x^8+8inBbb Q[x]$

I have this problem because I think the extension has degree $16$ but I can't decide the group: I think it could be $\Bbb Z_4\times\Bbb Z_2\times\Bbb Z_2$ but not surely:
First I do complex roots

$$x^8=-8=8e^{\pi i+2k\pi},\;k\in\Bbb Z\;,\;\;and\;then\;\;x_k=8^{1/8}e^{\frac\pi8\left(1+2k\right)},\;k=0,1,2,...,7$$

and I write $w=e^{\pi i/8}\;$ and this is primitive root of order $16$ , and then the roots are $8^{1/8}w^k,\;w=1,3,5,7,9,11,13,15$ .

Next I think maybe $8^{1/8}\in\Bbb Q(w)$ but this is not possible because $\;\Bbb Q(w)/\Bbb Q\;$ cyclotomic extension and its order $\phi(16)=8$, and $\;\Bbb Q(8^{1/8})/\Bbb Q\;$ also has the order $8$, and then we could get $\;\Bbb Q(8^{1/8})=\Bbb Q(w)\;$ , but the first is real field and the second has non real complex elements.

I had to do trigonometry but it only took me to $\,8^{1/4}\in\Bbb Q(w)\;$ because

$$\frac1{\sqrt2}=\cos\frac\pi4=2\cos^2\frac\pi8-1\implies\cos^2\frac\pi8=\frac{\frac1{\sqrt2}+1}2=\frac{1+\sqrt2}{2\sqrt2=\sqrt8}\implies\cos\frac\pi8=\frac{\sqrt{1+\sqrt2}}{\sqrt[4]8}$$

and I know also

$$w+w^{-1}=w+\overline w=2\,\text{Re}\,w=2\cos\frac\pi8\in\Bbb Q(w)$$

and this is why I get splitting field of degree $16$ :

$$\left[\Bbb Q(w,8^{1/8})\,:\,\Bbb Q\right]=\left[\Bbb Q(w)(8^{1/8}):\Bbb Q(w)\right]\left[\Bbb Q(w)\,:\,\Bbb Q\right]=2\cdot8=16$$

because $\;x^2-\sqrt[4]8\;$ is the minimal polynomial of $\;\sqrt[8]8\;$ over $\;\Bbb Q(w)\;$ .

Now, I know $\;\text{Gal}\,(\Bbb Q(w)\,/\,\Bbb Q)\cong\left(\Bbb Z/16\Bbb Z\right)^*\cong\Bbb Z_4\times\Bbb Z_2\;$ and is cyclotomic extension. Also, it is clearly $\;\Bbb Q(w,8^{1/8})/\Bbb Q\;$ cyclic extension of order two, and now questions

Question 1 Is there a simpler method to get $\;\sqrt[4]8\in\Bbb Q(w)\;$? It is not hard (only the trigonometry) but it is long and has many calculations and in time of exam perhaps one can use some lemma or theorem...?

Question 2 What is the $\;\text{Gal}\,(\Bbb Q(w,8^{1/8})\,/\,\Bbb Q)\,?$ Is it really $\;\Bbb Z_4\times \Bbb Z_2\times \Bbb Z_2\;$ or some harder group? I read there are 14 groups of order $16$, and only 5 of them are abelian, so is there some lemma or trick to know the group this time?

Question 3 Is my work above right? Any correction, suggestion or comments is very much appreciated, thank you.