The answer is Yes.
For the first question, suppose that $A$ is Dedekind infinite and let $f$ be an injection from $\omega$ into $A$. Then the function that maps $f(2n)$ to $f(n)$ for each $n\in\omega$, maps $f(2n+1)$ to $f(0)$ for each $n\in\omega$, and leaves all other elements of $A$ fixed, is a surjection from $A$ onto $A$ which is not Dedekind-finite-to-one.
For the second question, suppose that $A$ is dually Dedekind infinite and let $f$ be a surjection from $A$ onto $A\cup\{b\}$, where $b\notin A$. For each $n\in\omega$, let $A_n=(f^{n+1})^{-1}[\{b\}]$. Clearly, for all $m,n\in\omega$ with $m<n$, $f^{n-m}|A_n$ is a surjection from $A_n$ onto $A_m$. Then the function $g$ that maps each $x\in A_{2n}$ to $f^n(x)$ for all $n\in\omega$, maps each $x\in A_{2n+1}$ to a fixed $a\in A$ for all $n\in\omega$, and leaves all other elements of $A$ fixed, is a surjection from $A$ onto $A$ which is not dually-Dedekind-finite-to-one. $g$ is not dually-Dedekind-finite-to-one, because the inverse image of $a$ under $g$ includes $B=\bigcup_{n\in\omega}A_{2n+1}$, and $B$ is dually Dedekind infinite since $f^2|B$ is a surjection from $B$ onto $B\cup\{b\}$.