1) Let $G=$ Gal$(L/K)$, $p \in K$. Suppose $p$ is unramified and nonsplit. (Since only finitely many primes are ramified, it suffices to show that this cannot occur.) Since $p$ is unramified and nonsplit and $efg=|G|$, we see that $f=|G|$ and the decomposition group $D_p$ is isomorphic to $G$. But we also have that $D_p$ is isomorphic to the Galois group of the residue field of $L/K$ at $p$, which is cyclic of order $f$. This contradicts our hypothesis on $G$.
Let $\mathcal O_K$ be a Dedekind domain, $L$ the fraction field of $\mathcal O_K$. Let $L/K$ be a finite, separable extension, not necessarily Galois, of degree $p$. Let $N$ be the normal closure of $L/K$. Let $G=\text{Gal}(N/K)$ and $H=\text{Gal}(N/L)$.
Let $\mathfrak p$ be a prime of $K$ (i.e. of $\mathcal O_K$). Let $Q$ be a prime of $N$ above $\mathfrak p$. Let $G_Q$ denote the decomposition group of $Q$ over $K$. Then (as discussed in Neukirch pg. 55) there is a bijection from the set of double cosets $H\backslash G/G_Q$ to the set $P_\mathfrak p$ of primes of $L$ above $\mathfrak p$, given by: $$H\backslash G/G_Q\rightarrow P_\mathfrak p, \quad H\sigma G_Q\mapsto\sigma Q\cap L$$
Now suppose the prime $\mathfrak p$ is unramified in $L$. Then $\mathfrak p$ is also unramified in $N$.
Note that there are $p$ cosets $H\sigma_1,\dots,H\sigma_p$ of $H\backslash G$ where $p=[L:K]$. There is an action of $G$ that permutes the cosets $H\sigma_i$ by right multiplication. They key observation is this:
The size of the orbit of the coset $H\sigma_i$ under the right action of the decomposition group $G_Q$ equals the inertia degree of the prime $\sigma_iQ\cap L$ over $\mathfrak p$.
To show this, first observe that, for $\rho\in G_Q$ and $\sigma_i\in G$, $$H\sigma_i\rho=H\sigma_i \Longleftrightarrow \rho\in\sigma_i^{-1}H_{\sigma_iQ}\sigma_i$$ where $H_{\sigma_iQ}$ is the decomposition group of $\sigma_iQ$ over $L$.
Thus the size of the orbit of $H\sigma_i$ is $$[G_Q:\text{stab}(H\sigma_i)]=[G_Q:\sigma_i^{-1}H_{\sigma_iQ}\sigma_i]=[\sigma_iG_Q\sigma_i^{-1}:H_{\sigma_iQ}]=[G_{\sigma_iQ}:H_{\sigma_iQ}]$$
$[G_{\sigma_iQ}:H_{\sigma_iQ}]$ equals the inertia degree of $\sigma_iQ\cap L$ over $\mathfrak p$, proving the highlighted claim above.
Now assume the degree $p$ of $L/K$ is prime, and assume that $\mathfrak p$ has two prime factors $\mathfrak P_1$ and $\mathfrak P_2$ in $L$ of degree 1. This implies we have two cosets $H\sigma_1$ and $H\sigma_2$ who orbits under the action of $G_Q$ are of size 1. $G$ is a solvable group with a transitive action on the $p$ cosets $H\sigma_1,\dots,H\sigma_p$. Thus each element of $G_Q$ fixes the two cosets $H\sigma_1$ and $H\sigma_2$, so by the theorem given in the Hint, each element of $G_Q$ must fix all the cosets, so $G_Q$ partitions the $H\sigma_i$ into $p$ distinct orbits of one element each. Thus, every prime factor of $\mathfrak p$ in $L$ is of degree 1 over $\mathfrak p$.
Best Answer
It becomes easy if you use the preceding exercise in Neukirch's book:
The galois closure $N$ of $L|K$ is the composite field of the subfields $\sigma L$ where $\sigma$ ranges over $G(N|L)$. Since $\mathfrak p$ is totally split in $L$ it is also totally split in $\sigma L$ (if $\mathfrak p\mathcal O_L = \mathfrak P_1\cdots\mathfrak P_r$ in $L$ then $\mathfrak p \mathcal O_{\sigma L} = (\sigma \mathfrak P_1)\cdots(\sigma \mathfrak P_r)$ in $\sigma L$). So $\mathfrak p$ is also totally split in $\prod_{\sigma} \sigma L = N$ by the quoted exercise.