First let me answer the first question. Yes. It is possible. You can find the proof in Jech "The Axiom of Choice" in Theorems 11.2 and 11.3. Also you might be interested in reading Pincus' paper,

David Pincus, Cardinal representatives, Israel J. Math. 18 (1974), 321--344.

Where he shows that it is also consistent that there are representatives to all cardinals uniformly, but the axiom of choice fails.

To the third question, which seems to mix two separate ideas, I don't know the answer, and I don't know if it was asked in explicit details. But even without any assumptions on cardinal representatives, just talking about disjoint unions (and products), let me give you the following answer.

If you allow talking about products, then trivially you get the axiom of choice back.

Suppose that $\{Y_i\mid i\in I\}$ is a family of non-empty sets. Pick a large enough non-empty set $Y$ such that each $Y_i$ satisfies $|Y\times Y_i|=|Y|$.¹ Now consider $\prod_{i\in I} Y$ and $\prod_{i\in I}(Y\times Y_i)$, since the former is non-empty, the latter is non-empty either. Now pick some $f$ such that $f(i)\in Y\times Y_i$ and consider the projection onto $Y_i$, as a choice function.

For addition it seems to be open. The last progress I am aware of can be found in Higasikawa's paper,

Masasi Higasikawa, Partition principles and infinite sums of cardinal numbers. Notre Dame J. Formal Logic 36 (1995), no. 3, 425–434.

Where you can find a proof that this assumption implies The Partition Principle (whose choice strength is quite open for quite some time now). You can find other proofs related to this principle, and that in the presence of some additional assumptions it will imply the axiom of choice.

Two remarks that might be useful, taken from Gregory Moore's "Zermelo's Axiom of Choice":

1. The assumption is in fact equivalent to the statement that $\sum_{i\in I}|A|=|A|\cdot|I|$.

2. If we also assume that "If $Y$ is a set and for every $y\in Y$, $|y|=|Y|$, then $|\bigcup Y|=|Y|$, then we can prove the axiom of choice.

Footnotes.

1. Such $Y$ always exists. For example $(\bigcup Y_i)^\omega$.

First let me answer the first question. Yes. It is possible. You can find the proof in Jech "The Axiom of Choice" in Theorems 11.2 and 11.3. Also you might be interested in reading Pincus' paper,

David Pincus, Cardinal representatives, Israel J. Math. 18 (1974), 321--344.

Where he shows that it is also consistent that there are representatives to all cardinals uniformly, but the axiom of choice fails.

To the third question, which seems to mix two separate ideas, I don't know the answer, and I don't know if it was asked in explicit details. But even without any assumptions on cardinal representatives, just talking about disjoint unions (and products), let me give you the following answer.

If you allow talking about products, then trivially you get the axiom of choice back.

Suppose that $\{Y_i\mid i\in I\}$ is a family of non-empty sets. Pick a large enough non-empty set $Y$ such that each $Y_i$ satisfies $|Y\times Y_i|=|Y|$.¹ Now consider $\prod_{i\in I} Y$ and $\prod_{i\in I}(Y\times Y_i)$, since the former is non-empty, the latter is non-empty as well. Now pick some $f$ such that $f(i)\in Y\times Y_i$ and consider the projection onto $Y_i$, as a choice function.

For addition it seems to be open. The last progress I am aware of can be found in Higasikawa's paper,

Masasi Higasikawa, Partition principles and infinite sums of cardinal numbers. Notre Dame J. Formal Logic 36 (1995), no. 3, 425–434.

Where you can find a proof that this assumption implies The Partition Principle (whose choice strength is quite open for quite some time now). You can find other proofs related to this principle, and that in the presence of some additional assumptions it will imply the axiom of choice.

Two remarks that might be useful, taken from Gregory Moore's "Zermelo's Axiom of Choice":

1. The assumption is in fact equivalent to the statement that $\sum_{i\in I}|A|=|A|\cdot|I|$.

2. If we also assume that "If $Y$ is a set and for every $y\in Y$, $|y|=|Y|$, then $|\bigcup Y|=|Y|$, then we can prove the axiom of choice.

Footnotes.

1. Such $Y$ always exists. For example $(\bigcup Y_i)^\omega$.