Embedding of finite groups

Yes.

The symmetric group $Sym(n)$ is generated by $\{(1,2), (2,3),\ldots, (n−1,n)\}$. You can embed $Sym(n)$ into $Alt(n+2)$ as the group generated by $\{(1,2)(n+1,n+2), (2,3)(n+1,n+2), …, (n−1,n)(n+1,n+2)\}$. This embedding takes a permution $\pi\in Sym(n)$ and sends it to $\pi⋅(n+1,n+2)^{\text{sgn}(\pi)}$, where $\text{sgn}(\pi)\in\{0,1\}$ is the parity of the permutation.

In other words, $G\le Sym(n)\le Alt(n+2)$ embeds any group into a (slightly larger) alternating group.

The general linear group $GL(n,q)$ embeds in the special linear group $SL(n+1,q)$ using a determinant trick. We just add a new coordinate to cancel out the determinant of the matrix from $GL(n,q)$ so the result lands in $SL(n+1,q)$.

$$\operatorname{GL}(n,q) \cong \left\{ \begin{bmatrix} A & 0 \\ 0 & 1/\det(A) \end{bmatrix} : A \in \operatorname{GL}(n,q) \right\} ≤ \operatorname{SL}(n+1,q)$$

In other words, $G\le GL(n,q)\le SL(n+1,q)$ embeds any group into a (slightly larger) special linear group.

Yes we can.

For $A_n$, we can embed the given group in some $S_{n-2}$ and then for the additional two elements choose the identity or the transposition according as the element of $S_{n-2}$ is even or odd.

For $SL(n,q)$, we can embed the given group in some $GL(n-1,q)$ and then choose the diagonal element in the additional row and column as the reciprocal of the determinant of the element of $GL(n-1,q)$.

To your first question (embeddable within $A_n$) I think the answer is yes for obvious reasons: one can embed $S_n$ within $A_{n+2}$.

(Consider the subset of $A_{n+2}$ that stabilizes the first $n$ elements (as a set), it's obvious that this set will consist of all permutations of these elements, where it may or may not interchange the last two points, depending on the sign. For instance for $n=3$ we retrieve $S_3$ as $(1\ 2\ 3)$, $(1\ 2)(4\ 5)$, etc...)