For your question 1, you can use the fact that $\pi= \zeta_{p^n}-1$ is a root of the polynomial : $F(X)=(1+X)^{(p-1)p^{n-1}}+(1+X)^{(p-1)p^{n-2}} + \dots + 1$ which is an Eisenstein polynomial over $\mathbb{Q}_p$, and hence irreducible. But $\mathbb{Q}_p(\pi) = \mathbb{Q}_p(\zeta_{p^n}) = K$, so $[K : \mathbb{Q}_p] = \phi(n)$.
On the other side, to show that $K/\mathbb{Q}_p$ is totally ramified, you can notice that $F(0) = p = \prod_{\sigma \in Gal(K/\mathbb{Q}_p)} \sigma(\pi)$. But we have also : $|\pi|_K = |\sigma(\pi)|_K$, hence $|p|_K = |\pi^{\phi(n)}|_K$. So, we have : $p \in \pi^{\phi(n)}O_K^{\times}$. But as $\phi(n)$ is also the degree of $K/\mathbb{Q}_p$, you can conclude directly that $\pi$ is a uniformizer and $K/ \mathbb{Q}_p$ is totally ramified. Moreover, we can deduce from everything above that the conjuguate of $\zeta_{p^n}$ is the $(\zeta_{p^n}^{i})_{gcd(i,p) = 1, i \in \{ 0, \dots, p^n-1 \}}$, and then the Galois group of $K/\mathbb{Q}_p$ is clearly isomorphic to $(\mathbb{Z}/p^n\mathbb{Z})^{\times}$.
Finally, to find $O_K$, you can prove that $O_K \cap \mathbb{Q}_p[\pi] = \mathbb{Z}_p[\pi]$.
Let $K$ be a local field with residue field $k$ and uniformiser $\pi$. Let $F$ be a finite extension with residue field $k_F$ and uniformiser $\pi_F$. An extension $F/K$ can take one of two flavours:
If $\pi\mathcal O_F = (\pi_F)^e$, we say the extension is ramified. It is totally ramified if $e = [F:K]$.
If $\pi\mathcal O_F = (\pi_F)$ is still prime, we say the extension is unramified. In this case, we can choose $\pi_F = \pi$.
In the ramified case, the extension is obtained by adding a new uniformiser of smaller valuation: if we normalise the valuation on $F$ so that $v(\pi_F) = 1$, then $v(\pi) = e$. In the totally ramified case, the residue field $k_F = k$.
In the unramified case, the uniformiser does not change. Instead, the extension is happening on the residue field: $[k_F: k] = [F:K]$.
Now, suppose that $F = K(\alpha)$ where $\alpha$ is a root of the irreducible polynomial $f$. We can always assume that $f$ is monic with integer coefficients. In this case, $\alpha$ is an algebraic integer. Let $\overline \alpha$ be its image in $k_F$. Then $\overline \alpha$ is a root of $f\pmod \pi$. In particular, if $F/K$ is totally ramified, then $k(\overline \alpha) = k$. It follows that $f\pmod \pi$ is completely reducible.
The converse is false in general, as $f$ could be a polynomial like $X^2 + p^2$, and $F(\sqrt{-p^2}) = F(\sqrt{-1})$ is unramified if the residue characteristic is not $2$. The problem here is that the map $\mathcal O_K[\sqrt{-p^2}]\to k_F$ is not surjective: its image is exactly $k$! However, if $\mathcal O_F = \mathcal O_K[\alpha]$, then the map $\mathcal O_K[\alpha]\to k_F$ is surjective. In particular, $k_F = k(\overline \alpha)$. So if $f\pmod \pi$ is completely reducible, then $k_F = k$, so $F$ is totally ramified.
Best Answer
If $K$ is a global field, $p$ is a prime of $\mathcal{O}_K$, and $L/K$ is an extension with $n=[L:K]$, we say that $p$ is totally ramified in $L$ (or that $L$ is totally ramified at $p$) if $p\mathcal{O}_L=q^n$ for some prime $q$ of $L$.
However, since of course there are many primes of $K$, it's not a complete statement to only say "The field $L$ is totally ramified over $K$", you have to specify which prime you're talking about – alternatively, the statement could just mean "There exists at least one prime $p$ of $K$ that is totally ramified in $L$".