Consider the map $\phi:(F^*,.) \to SL_2(F)$ defined as $\phi(a)= \begin{pmatrix}a& 0 \\0& a^{-1} \end{pmatrix} $ . We have $\phi(1)=\begin{pmatrix} 1 &0 \\ 0&1\end{pmatrix}$ and for $a, b\in F^*$, $\phi(ab)=\begin{pmatrix}ab & 0 \\ 0 & (ab)^{-1}\end{pmatrix}=\begin{pmatrix} a &0 \\ 0 & a^{-1} \end{pmatrix} \begin{pmatrix} b & 0 \\0 &b^{-1} \end{pmatrix}=\phi(a)\phi(b)$. So it is indeed a group homomorphism.
Now consider $\phi(a) = \phi(b) \implies\begin{pmatrix} a &0 \\ 0 & a^{-1} \end{pmatrix}= \begin{pmatrix} b & 0 \\0 &b^{-1} \end{pmatrix}\implies a = b $ and $a^{-1}=b^{-1}$, i.e $\phi$ is injective. So $\phi$ does embed $(F^*,.)$ into $SL_2(F)$.
For the embedding of $(F,+)$ into $SL_2(F)$, consider the map $\psi(x)=\begin{pmatrix}1 & x \\ 0 & 1\end{pmatrix}$. We have $\psi(x)\psi(y)=\begin{pmatrix}1 & x \\ 0 & 1\end{pmatrix}\begin{pmatrix}1 & y \\ 0 & 1\end{pmatrix}=\begin{pmatrix}1 & x+y \\ 0 & 1\end{pmatrix}=\psi(x+y)$, therefore it is a group homomorphism.
Now, $\psi(x)=\psi(y)\implies \begin{pmatrix}1 & x \\ 0 & 1\end{pmatrix} =\begin{pmatrix}1 & y \\ 0 & 1\end{pmatrix}\implies x=y$, therefore $\psi(x)=\psi(y)\implies x=y$.
Thereby we conclude that $\psi:(F,+) \to SL_2(F)$ is in fact an embedding.
Edit:
Initially I used: $(F,+)$ into $SL_2(F)$, we consider the map $\psi:(F,+) \to SL_2(F)$ defined as $\psi(a)=\begin{pmatrix}\cos(a) & -\sin(a) \\\sin(a)& \ \ \ \cos(a)\end{pmatrix}$.
which does not make sense as pointed out in the answers since the domain of these functions is the real line. Even when $F=\mathbb{R}$ it isn't an embedding since it is not one-one.
Kindly verify.
Your answer to the second part makes no sense, because there are no $\sin$ or $\cos$ functions defined on a general field. Even for $F=\Bbb R$, it is not an embedding, as the map is not injective.
Instead, try looking at matrices of the form $\begin{pmatrix} 1 & x\\0& 1\end{pmatrix}$. These are known as unipotent matrices.