Is it possible for some non-Hermitian matrix $A \in M_n({\mathbb{C}})$ to satisfy
$1)$ $x^t Ax \ge 0 $ for all $x\in \mathbb{R^n}$ ?
$2)$ $x^* Ax \ge 0 $ for all $x \in \mathbb{C^n}$ ?
i take $ A =\begin{bmatrix} 0 &1 \\ 2& 0 \end{bmatrix}$, i thinks both
$1$ and $2$ possible ...
Real positive semidefinite matrices
There exist non-symmetric real matrices that are positive semidefinite. For example, take
$$A=\begin{pmatrix} 1 & 1\\ -1 & 1 \end{pmatrix}.$$
Then, we get $x^T A x=x_1^2+x_2^2\geq 0\,$ for all $x\in \mathbb{R}^2$. Your example does not work, since $$x^T\begin{pmatrix} 0 &1 \\ 2& 0 \end{pmatrix}x=3x_1x_2,$$ which is not non-negative for all $x\in\mathbb{R}^2$.
It is useful to look at symmetric positive semidefinite matrices because you get nice information about the eigenvalues, spectral decomposition, and so on. But even for non-symmetric positive semidefinite matrices you can say something about the eigenvalues. We know that a symmetric positive semidefinite matrix has only non-negative eigenvalues. A non-symmetric positive semidefinite matrix can have complex eigenvalues and therefore, saying that a complex eigenvalue is positive does not make any sense. But you can show that the real parts of the eigenvalues of a positive semidefinite (not necessarily symmetric) real matrix are non-negative. For example, the matrix $A$ has the eigenvalues $1\pm i$.
Complex positive semidefinite matrices
You can show that every complex positive semidefinite matrix is Hermitian. See, e.g., this post Positive definite matrix must be Hermitian.
