The answer to the first question depends on what you mean by "has".
The rotation of $90^{\circ}$ about the origin in $\mathbb{R}^2$ has no eigenvalues when working over $\mathbb{R}$: $\left(\begin{array}{rr} 0 & -1\\ 1 & 0 \end{array}\right)$ while its square has one (repeated) eigenvalue, namely $-1$. This phenomenon is common when working over fields that are not algebraically closed.
However, if we consider the eigenvalues to be complex numbers (even though the original matrix has coefficients in $\mathbb{R}$), then the above is not a counterexample, since the roots of the characteristic polynomial of the rotation are $i$ and $-i$ (two distinct), while the roots of the characteristic polynomial of the square is $-1$ doubled (one root).
So let's ask:
Can the characteristic polynomial of $A^2$ have more distinct roots (in the algebraic closure of the ground field) than the characteristic polynomial of $A$?
The answer is "no."
Over an algebraically closd field every matrix can be put in upper triangular form (e.g., via the Jordan canonical form). Thus, we may assume that $A$ is of the form $A=\left(\begin{array}{cccc} \lambda_1 & * & \cdots & *\\ 0 & \lambda_2 & \cdots & *\\ 0 & 0 & \cdots & *\\ \vdots & \vdots &\ddots & \vdots\\ 0 & 0 & \cdots & \lambda_n \end{array}\right).$ Therefore, $A^2 = \left(\begin{array}{cccc} \lambda_1^2 & \# & \cdots & \#\\ 0 & \lambda_2^2 &\cdots & \#\\ 0 & 0 & \cdots & \#\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & \lambda_n^2 \end{array}\right)$ and the set $\{\lambda_1^2,\ldots,\lambda_n^2\}$ (the roots of the characteristic polynomial of $A^2$) contains no more distinct elements than $\{\lambda_1,\ldots,\lambda_n\}$ (the roots of the characteristic polynomial of $A$), and possibly fewer.