Does every ideal class contains a prime ideal that splits?
Yes, every ideal class contains a split prime. Here is one possible proof, which however doesn't satisfy the conditions of part B of your question: the analogue of Dirichlet's argument, using ideal class characters, shows that $\sum_{\mathfrak p \in [I]} N\mathfrak p^{-1}$ (the sum being taken over prime ideals in the class of $I$) diverges. On the other hand, the sum over non-split primes converges (because a non-split prime lying over $p$ has norm at least $p^2$, and so the sum over non-split primes is majorized by (some constant times) $\sum_p p^{-2}$, which converges). Thus there must be infinitely many split primes contributing to this sum.
Note: As the OP points out in a comment below, this argument applies only in the Galois case, and so doesn't actually address the question.
The answer to both A and B is no. Take the non-Galois field $L = \mathbb{Q} (\sqrt[4]{-5})$, and let $N = L(i)$, the Galois closure of $L$. Computing with PARI/GP tells me $L$ has cyclic class group of order $4$ and $N$ has class group of order $2$. Thus, the map $N \colon \mathrm{Cl}_N \rightarrow \mathrm{Cl}_L$ induced by the relative norm of ideals is not surjective.
Assume, to reach a contradiction, that one of the classes of order $4$ in $\mathrm{Cl}_L$ contains a prime $\mathfrak{p}$ lying over a prime $p \in \mathbb{Z}$ that is completely split in $L$. It is well-known that this happens if and only if $p$ is completely split in $N$. Then $\mathfrak{p}$ is the norm of an ideal $\mathfrak{P}$ in $N$. But then the class of $\mathfrak{p}$ is the norm of the class of $\mathfrak{P}$, contradicting the fact that the norm map on classes is not surjective.
A is true if we make the additional assumption that the normal closure of $L$ is linearly disjoint over $L$ from the Hilbert class field of $L$.
To see this, let $N$ be the normal closure of $L$ over $\mathbb{Q}$, let $H_N$ be the Hilbert class field of $N$, and let $H_L$ be the Hilbert class field of $L$. We must show that for each $\sigma \in G(H_L/L)$, there exists some prime ideal of $L$, completely split over $\mathbb{Q}$ and whose Frobenius for $H_L/L$ is $\sigma$.
We need some facts:
- $H_N$ is Galois over $\mathbb{Q}$
- $N H_L \subset H_N$
For 1, consider an embedding $\tau \colon H_N \hookrightarrow \overline{\mathbb{Q}}$. We have $\tau (N) = N$, so $\tau (H_N)$ is an abelian unramified extension of $N$, hence is contained in $H_N$ It follows that $\tau(H_N) = H_N$ and 1 follows.
For 2, consider a prime ideal $\tilde{\mathfrak{P}}$ in $N H_L$, which is Galois over $L$. Because $H_L/L$ is unramified, the inertia group for $\tilde{\mathfrak{P}}$ over $L$ is contained in $G(N H_L /H_L)$. But the definition of inertia groups also shows that the inertia group for $\tilde{\mathfrak{P}}$ over $N$ is contained in that for $\tilde{\mathfrak{P}}$ over $L$. Thus, the inertia group for $\tilde{\mathfrak{P}}$ over $N$ consists only of automorphisms that fix $N$ and $H_L$, so it is trivial and $N H_L/N$ is unramified. Since $G(N H_L/N)$ injects into $G(H_L/L)$ by restriction, $N H_L/N$ is abelian and 2 follows.
Now let us return to considering our $\sigma$ in $G(H_L/L)$. By 1 and the assumption that $N$ and $H_L$ are linearly disjoint over $L$, we can lift $\sigma$ to an automorphism $\tilde{\sigma}$ in $G(H_N/N)$. By Chebotarev's density theorem, we can find some prime ideal $\mathfrak{P}$ in $H_N$, unramified over $\mathbb{Q}$ and whose Frobenius over $\mathbb{Q}$ is $\tilde{\sigma}$. The decomposition group of $\mathfrak{P}$ over $\mathbb{Q}$ is thus contained in $G(H_L/L)$, so the prime $p = \mathfrak{P} \cap \mathbb{Z}$ is completely split in $L$.
Let $\mathfrak{p} = \mathfrak{P} \cap N$. By the formalism of the Artin symbol, the Frobenius for $\mathfrak{P}$ over $N$ is $\tilde{\sigma}$ and its restriction to $H_L$, namely $\sigma$ is itself the image under the Artin map of the ideal $N_{N/L} \mathfrak{p}$. But again, because $p$ is completely split up to $L$, it is also completely split up to $N$ and so $N_{N/L} \mathfrak{p}$ is a prime ideal of $L$ with degree $1$ and Frobenius equal to $\sigma$ as desired.