There is no largest natural number. The reason is that by Cantor's theorem, the power set of a finite set is a strictly larger set, and one can prove inductively that the power set of a finite set is still finite.

All numbers of the form $2^n$ for natural numbers $n\geq 1$ are even. The reason is that the power set of an $n$-element set has size $2^n$, proved by induction, and this is a Boolean algebra, which can be decomposed into complementary pairs $\{a,\neg a\}$. So it is a multiple of $2$.

Every finite set can be well-ordered. This follows by the Axiom of Choice via the Well-order Principle, which asserts that every set can be well-ordered.

Every non-empty set $A$ has at least one element. The reason is that if $A$ is nonempty, then $\{A\}$ is a family of nonempty sets, and so by the Axiom of Choice it admits a choice function $f$, which selects an element $f(A)\in A$.

There is no largest natural number. The reason is that by Cantor's theorem, the power set of a finite set is a strictly larger set, and one can prove inductively that the power set of a finite set is still finite.

All numbers of the form $2^n$ for natural numbers $n\geq 1$ are even. The reason is that the power set of an $n$-element set has size $2^n$, proved by induction, and this is a Boolean algebra, which can be decomposed into complementary pairs $\{a,\neg a\}$. So it is a multiple of $2$.

Every finite set can be well-ordered. This follows by the Axiom of Choice via the Well-ordering Principle, which asserts that every set can be well-ordered.

Every non-empty set $A$ has at least one element. The reason is that if $A$ is nonempty, then $\{A\}$ is a family of nonempty sets, and so by the Axiom of Choice it admits a choice function $f$, which selects an element $f(A)\in A$.