$\def\Q{\mathbf{Q}}$
$\def\Z{\mathbf{Z}}$
$\def\Br{\mathrm{Br}}$
$\def\inv{\mathrm{inv}}$
$\def\Gal{\mathrm{Gal}}$
The questions you ask are essentially straightforward enough, but require a little theory. In fact, I once asked a starting graduate student to answer your question as preparation for his qualifying exam. (Hint: if you ever tell your advisor "I basically understand the statements of class field theory but I don't really know anything about division algebras" be prepared for the answer "OK, your qual will center on explaining CFT through Brauer groups.") I guess he never came back to this website to actually post an answer to this question, and I'm bored now, so here's an answer.
I'm going to begin by stating a bunch of very standard results without either proofs or references. Any standard text on class field theory (including Milne's online notes) will have all the details.
Lemma: If $K/\Q$ can be centrally embedded into $D/\Q$ of dimension $n^2$, then there is a finite extension $L/K$ where $L$ can be centrally embedded into $D/\Q$ and $[L:\Q] = n$.
This is another way of saying that all maximal commutative subfields have degree $n$.
Lemma: $L/\Q$ of dimension $n$ can be centrally embedded into $D/\Q$ of dimension $n^2$ if and only if $L$ splits $D$.
Standard.
Theorem: There is an exact sequence:
$$0 \rightarrow \Br(\Q) \rightarrow \bigoplus_{p,\infty} \Br(\Q_p) \rightarrow \Q/\Z \rightarrow 0.$$
This is a key result of global class field theory.
Lemma: $\Br(\Q_p) = \Q/\Z$, $\Br(\mathbf{R}) = \Z/2\Z$.
This follows from local class field theory.
Lemma: Let $L_v/\Q_p$ be a finite extension. Then $D_p/\Q_p$ splits over $L_v$ if and only if $[L_v:\Q_p]$ annihilates the class $D_p \in \Br(\Q_p)$, or equivalently
$$[L_v:\Q_p] \inv(D) = 0,$$
where $\inv_p(D) \in \Q/\Z$ or $\Z/2\Z$ depending on whether $p$ is finite or not.
Standard.
Now, simply by connecting the dots, one obtains the following:
Lemma: A field $L/\Q$ of degree $n$ embeds centrally and maximally into $D/\Q$ of dimension $n^2$ if and only if,
for all primes $p$ and for all $v|p$ in $p$,
$$[L_v:\Q_p] \inv_p(D) = 0.$$
There is a lot of freedom in constructing $D$. We can make $\inv_p(D)$ anything we want, as long as it is zero for all but finitely many primes $p$, has order dividing $2$ at infinity, and satisfies
$$\sum_p \inv_p(D) = 0.$$
Hence we also deduce:
Lemma: A field $L/\Q$ of degree $n$ embeds centrally and maximally into $D/\Q$ of dimension $n^2$ if and only if,
for all primes $q | n$ with $q^m \| n$, we can find two distinct primes $p_i$ for $i = 1,2$ (one possibly infinite) such that, for any
prime $v|p_i$, we have
$$q^m | [L_v:\Q_{p_i}].$$
Proof: Since $\sum \inv_p(D) = 0$ and $D$ has order $n$, there exist at least two (possibly one is infinite if $q^m = 2$) primes $p_i$ such that $\inv_p(D)$ has order divisible by $q^m$. This gives one direction.
For the converse, take the invariants to be the sum of the vector over $q^m | n$ whose contribution is $1/q^m$ at $p_1$ and $-1/q^m$ at $p_2$ and zero elsewhere.
Now we deduce:
Claim: A field $K/\Q$ of degree $n$ embeds into $D/\Q$ of dimension $N^2$ for some $N$ if and only if,
for all primes $q | n$ with $q^m \| n$, we can find two distinct primes $p_i$ (one possibly infinite) such that, for any
prime $v|p_i$, we have
$$q^m | [L_v:\Q_{p_i}].$$
Proof: If there is such an embedding, there is a maximal embedding of an $L$ which contains $K$.
Suppose that $N = dn$, and let $q^M \| d$ so $q^{m+M} \| N$. We deduce from the previous Lemma that there is a prime $p_i$ such that
$$q^{m+M} | [L_w:\Q_{p_i}] = [L_w:K_v][K_v:\Q_{p_i}]$$
for all $v|p_i$ and all $w|v$. Hence it follows that
$$q^{m+M} | \sum_{w|v} [L_w:K_v] [K_v:\Q_{p_i}] = [L:K][K_v:\Q_{p_i}] = d [K_v:\Q_{p_i}],$$
and thus
$$q^m | [K_v:\Q_{p_i}],$$
as required. For the converse, one can easily construct a totally ramified $L/K$ which has $L_w = K_v$
for all $v|p_i$ for any finite set of $p_i$, and then use the previous Lemma.
The claim basically gives an answer to your first question, note that one also deduces that $K$ is included inside some $D$ if and only if it is included maximally for some $D$ of dimension $[K:\Q]^2$. The second question presents no real extra difficulties.
Let me give some examples.
Example: $K/\Q$ has prime degree $q$. Let $L$ be the Galois closure of $K$. We have $G = \Gal(L/\Q) \subset S_q$ has order divisible by $q$, and hence contains a $q$-cycle. Thus by Chebotarev there are infinitely many $p_i$ such that Frobenius at $p_i$ is a $q$-cycle. This implies that $[K_v:\Q_{p_i}] = q$ for all $v|q$, and so $K$ embeds.
Example: $K/\Q$ is Galois and cyclic. In this case, $G = \Gal(K/\Q) = \Z/n \Z \subset S_n$ also has an $n$ cycle, so as in the last question one is done by Cebotarev, and $K$ embeds.
Example: $K/\Q$ is bi-quadratic, and $[K_v:\Q_p] \le 2$ for all ramified primes. In this case, $K$ can never be embedded inside a $D/\Q$, because there are no primes $p_i$ such that $4 | [K_v:\Q_p]$.
Remark In the last example, one does require the hypothesis at the ramified primes, since it could happen that $[K_v:\Q_p] = 4$ for two primes dividing $\Delta_K$. Thus the answer depends not only on the Galois group but also at the inertia groups at primes of bad reduction.
Example: $K = \Q(\sqrt{-1},\sqrt{17})$ does not embed centrally into any $D/\Q$. This is a special case of the last example.
Example: $K = \Q(\sqrt{-1},\sqrt{3})$ does embed centrally into a $D/\Q$ (for example with invariants $\inv_2(D) = 1/4$ and $\inv_3(D) = -1/4$ and $\inv_p(D) = 0$ otherwise). This is a special non-case of the last example, because $[K_3:\Q_3] = 4$ and $[K_2:\Q_2] = 4$.
Example: $K/\Q$ is totally ramified at two primes $p_1$ and $p_2$. In this case, $[K_v:\Q_{p_i}] = [K:\Q]$, so we can aways embed $K$.
I think these examples are enough to get the point.
