Extension fields isomorphic to fields of matrices

The answer to your first question is yes. In particular, if $L/K$ has degree $m$ then $L$ is isomorphic to a subfield of $M_{m \times m}(K)$. This can be seen via the (left) regular representation as follows.

Each $\alpha \in L$ gives rise to a (left) multiplication map $\lambda_\alpha : L \to L$, $x \mapsto \alpha x$ that is $K$-linear. Choosing a basis of $L$ as a $K$-vector space allows us to write a matrix representation of $\lambda_\alpha$ with respect to this basis. For instance, in your example $\mathbb{C}$ has basis $\{1, i\}$ as an $\mathbb{R}$-vector space. Given $\alpha = a + ib \in \mathbb{C}$, then \begin{align*} \lambda_\alpha(1) &= \alpha = a + ib\\ \lambda_\alpha(i) &= \alpha \cdot i = (a + ib)i = -b + ia \end{align*} so the matrix representation of $\lambda_\alpha$ with respect to this basis is $\begin{pmatrix} a & -b\\ b & a \end{pmatrix}$.

Here's a partial answer to your second question. Suppose $L/K$ has a primitive element, so $L = K(\beta)$ for some $\beta \in L$. (This is true for instance if $L/K$ is a separable extension.) Let $n$ be the minimal positive integer such that $L$ embeds into $M_{n \times n}(K)$; denote this embedding by $\iota : L \to M_{n \times n}(K)$. One can show that $\beta$ satisfies the characteristic polynomial $f$ of $\iota(\beta)$, which has degree $n$. Since $\beta$ generates the extension $L/K$ and $[L:K] = m$, then the minimal polynomial $g$ of $\beta$ over $K$ has degree $m$. Since $\beta$ is also a root of $f$, then $g \mid f$, so $m \leq n$. Thus $m=n$.

I don't know the answer to your second question in general. Maybe someone with more background in representation theory will have an answer.