Perhaps lets start with the relations between surjections, entire relations, and indexed families of nonempty sets.
It turns out these three notions are equivalent.
Suppose $f:A\to B$ is a surjection. Define a relation $R:B\to A$ by $bRa\iff f(a)=b$. Since $f$ is surjective, this relation is entire. On the other hand, if $b\in B$, define $A_b = f^{-1}(\{b\})$, since $f$ is a surjection, each $A_b$ is nonempty, so we have a family of (disjoint) nonempty sets indexed by $B$.
Now suppose we have an entire relation $R:B\to A$. Define $A_b = \{a\in A: bRa\}$, which gives a family of nonempty sets indexed by $B$, since $R$ is entire. Finally, define $$A'=\bigsqcup_{b\in B} A_b,$$
and $f:A'\to B$ by $f(a,b)=b$.
Lastly, suppose we start with a family of nonempty sets indexed by $B$, $A_b$. Then again, we define $A'=\bigsqcup_{b\in B} A_b$, and $f:A'\to B$ by $f(a,b)=b$, which is surjective, since all the $A_b$ are nonempty. On the other hand, we can define an entire relation $R:B\to A'$ by $b R (a,b)$. (Or we could take $A=\bigcup_{b\in B} A_b$ and $R:B\to A$ by $bRa \iff a\in A_b$.)
Choice
One version of the axiom of choice says that if $A_b$ is a family of nonempty sets indexed by $B$, then there is a function
$$g:B\to A'= \bigsqcup_{b\in B} A_b$$
such that $fg=1_B$, where $f:A'\to B$ is the surjective function constructed above.
$g$ is called a choice function.
Now the relationship between the statements of choice in the question is the following:
The following are equivalent
1. Choice (as stated just now)
2. Every surjective function has a right inverse.
3. Every set is projective
4. Every entire relation contains a function
Proof
(1) $\implies$ (2): Given a surjective function $f:A\to B$, and applying choice to the family of sets $A_b=f^{-1}(b)$, we get a function $g:B\to A$ such that $fg =1_B$.
(2) $\implies$ (3): Suppose $f:A\to B$ is surjective, and $h:X\to B$ is any map of sets.
To show that all sets are projective, it suffices to show that we can always lift $h$ to a map $\tilde{h}:X\to A$. However, if $g:B\to A$ is a left inverse, then we can take $\tilde{h}= gh$, since then $f\tilde{h}=fgh=h$.
(3) $\implies$ (1): Suppose $A_b$ is a family of nonempty sets. Then $f : A'\to B$ is surjective, and $B$ is projective, so we can lift $1_B$ along $f$ to a map $g:B\to A'$ such that $fg=1_B$, which is the statement of choice.
(4) $\implies$ (2): If $f:A\to B$ is surjective, and $R:B\to A$ is the entire relation constructed above, and $g:B\to A$ is a function contained in $R$, then by definition,
$bRg(b)$, which means that $fg(b)=b$, so $g$ is a right inverse to $f$.
(1) $\implies$ (4): If $R : B\to A$ is an entire relation, then we defined a family of nonempty subsets $A_b=\{a\in A: bRa\}$. Letting $\tilde{g}:B\to A'$ be a choice function for this family, we have $\tilde{g}(b) = (a,b)$ for some $a$ with $bRa$, and we define $g:B\to A$ by $g(b)=a$, which gives a function contained in $R$.
$\blacksquare$
The relations of the statements in your question
(1) is the definition of projective, which is used in statement (5).
I just showed (2) and (4) are equivalent to choice.
(6) is equivalent to saying that any family of nonempty sets indexed by $B$ has a choice function, so it's choice for sets indexed by that set.
Best Answer
Yes. If you have a countable family of sets, $\{X_n\mid n\in\Bbb N\}$, extend it to a family of size $\Bbb R$.
Yes. "Every set of size $\Bbb R$ admits a choice function".
The Axiom of Choice, and therefore all of its equivalents: e.g. every vector space has a basis; every commutative unital ring has a maximal ideal; etc. Including every equivalent of the Boolean Prime Ideal theorem (e.g. every commutative unital ring has a prime ideal).
This is not particularly clear, but likely to be false.
This is Form 181 in the Howard–Rubin dictionary of choice principles. But the biggest problem with this principle is that it will follow from the conjunction of:
And while it's a somewhat nontrivial conjunction, it is also not particularly powerful or interesting (for example, it is easy to arrange a model where this conjunction holds, but there are sets which cannot be linearly ordered). You can find more information in the book, or in the online graphing website: https://cgraph.inters.co/.