Let $\mathbb{Q}\_{p}$ be a p-adic field such that $ p \neq 2 $. We knew that for every $ n=2m $ there exists exactly one unramified extension $ K $ of $ \mathbb{Q}\_{p} $ of degree $ n $, obtained by adjoining $ (p^n -1)$-th roots of unity to $ \mathbb{Q}\_{p}$
. Moreover $ Gal(K / \mathbb{Q}\_p) $ is isomorphic to $Gal(F\_{p^n} : F\_p) $ which is cyclic and generated by the Frobenius map.
Then what will be the generator of $ Gal(K / \mathbb{Q}\_p) $???
Moreover if $ p\equiv 3 (mod 4 ) $ then does primitive $ 8^th $ root of unity belonging to the the quadratic extension in the extension $(K / \mathbb{Q}\_p) $?



Unramified extension over $ \mathbb{Q}_{p} $
You did not specify $f$: it should be equal to $n$. A particular generator of the resulting cyclic Galois group is the Frobenius automorphism, which acts on the $(p^n−1)$-th roots of unity by $x\mapsto x^p$. The quadratic subextension is generated by the $(p^2-1)$-th roots of unity. As $p^2-1$ is divisible by $8$, this subextension contains a primitive $8$-th root of unity.


