The set of all countably-infinite subsets of an infinite set Let $A$ be an infinite set and $D(A)$ denote the set of all countably-infinite subsets of $A$ and let $P(A)$ denote ,as usual, the power set of $A$, then 
(i) does there exist a surjection of $D(A)$ onto $P(A)$ ?
(ii) if $A$ is uncountable then does there exist an injection from $A$ into $D(A)$ ?
(iii) if answer to (i) is "no" , then does there exist a surjection of $A$ onto $D(A)$ ? 
( I only know that $D(A)$ is always uncountable)
 A: The answers to 1 and 3 are sometimes; whereas the answer to 2 is trivially yes (even if $A$ is countable).
Let me start with 2, as it's a simple positive answer. Pick $X\subseteq A$ which is countably infinite, then consider the function: $$f(a)=\begin{cases} X\cup\{a\} & a\notin X\\ X\setminus\{a\} & a\in X\end{cases}$$
You can show that this is an injection (note that we didn't even care that $A$ is uncountable, just infinite).
The answer to (1) is easily no in some cases. For example if $A=\Bbb R$ then the set of countably infinite subsets of $\Bbb R$ has the same cardinality of $\Bbb R$. Therefore there is no surjection from $D(\Bbb R)$ onto $\mathcal P(\Bbb R)$; and there is certainly a surjection from $\Bbb R$ onto $D(\Bbb R)$.
However it is possible that the answer is yes. Consider the case where $\sf GCH$ holds, in that case if $|A|=\aleph_\omega$ we have that $|D(A)|=|\mathcal P(A)|$, in which case 3 is irrelevant.
But it could be the case that a statement known as the Singular Cardinal Hypothesis, or $\sf SCH$, fails. This statement follows from $\sf GCH$, and it is possible to have, for example $A$ whose cardinality is $\aleph_\omega$ as before, but all the following inequalities are sharp: $$|A|<|D(A)|<|\mathcal P(A)|$$
The consistency of this failure is stronger than that of $\sf ZFC$, and requires large cardinals.

Here is another example for the failure of 1 and 3 without using large cardinals. 
Consider a model in which $2^{\aleph_0}=\aleph_2$ and $2^{\aleph_1}=\aleph_3$. Let $A$ be a set of size $\aleph_1$. Then one can show that $|D(A)|=\aleph_2$, and now we have that $$|A|<|D(A)|<|\mathcal P(A)|$$ without resorting to very difficult results and large cardinals.
