Suppose you have two categories C and D, and functors $F:C\to D$ and $G:D\to C$ such that for all $x\in C$, $G(F(x))$ is isomorphic to $x$, and for all $y\in D$, $F(G(y))$ is isomorphic to $y$. If you didn't know how important it was to distinguish between "isomorphic" and "an isomorphism," you might think that then C and D are essentially the same category. But of course in order for C and D to be equivalent, one additionally needs the isomorphisms in question to be natural.
A nice example of a pair of functor with the above properties, but which are not an equivalence of categories, are the functors relating vector spaces and affine spaces. Here $F:\mathrm{Vect}\to \mathrm{Aff}$ regards a vector space as an affine space by forgetting its origin, and $G:\mathrm{Aff}\to \mathrm{Vect}$ constructs the "vector space of displacements" in an affine space. The composite $G F$ is naturally isomorphic to the identity, but $F G$ is only "unnaturally" isomorphic to the identity, and the categories are definitely not equivalent. A simpler version of this example relates groups with heaps.
One of my favourite examples is the following theorem, due to S. Mori:
Theorem A. Let $X$ be a smooth complex projective variety such that $-K_X$ is ample. Then $X$ contains a rational curve. In fact, through any point $x \in X$ there is a rational curve $D$ such that $$ 0 < -(D \cdot K_X )\leq \dim X+1.$$
In other words, smooth Fano varieties over $\mathbb{C}$ are uniruled.
The proof of this beautiful result uses deformation theory in a very striking way. The idea is the following. One first take any map $f \colon C \to X$, where $C$ is a smooth curve with a marked point $0$ such that $f(0)=x$.
Now by deformation theory of maps one knows that, if one requires that the image of $0 \in C$ is fixed, the morphism $f$ has a deformation space of dimension at least
$$h^0(C, f^*T_X)-h^1(C, f^*T_X) - \dim X = -((f_*C) \cdot K_X)-g(C) \cdot \dim X.$$
So, whenever the quantity $-((f_*C) \cdot K_X)-g(C) \cdot \dim X$ is positive, there must be a non-trivial family of deformations of the map $f \colon C \to X$ keeping the image of $0$ fixed. Then, by another result of Mori known as bend and break, one is able to show that at some point the image curve splits in several components and that one of them is necessarily a rational curve passing through $x$.
Instead, when $-((f_*C) \cdot K_X)-g(C) \cdot \dim X$ is not positive we are in trouble. But here comes another brilliant idea of Mori: let's pass to positive characteristic! In fact, in positive characteristic we may compose $f \colon C \to X$ with (some power of) the Frobenius endomorphism $F_p \colon C \to C$. This increases the quantity $-((f_*C) \cdot K_X)$ without changing $g(C)$ and allows us to obtain a deformation space which has again strictly positive dimension. So, using the argument above (deformation theory of maps + bend and break), for any prime integer $p$ we are able to find a rational curve through $x_p \in X_p$, where $X_p$ is the reduction of $X$ modulo $p$ (for the sake of simplicity I'm assuming that $X$ is defined over the integers).
Finally, a straightforward argument using elimination theory shows that if $X_p$ admits a rational curve through $x_p$ for every prime $p$, then $X$ admits a rational curve through $x$, too.
It is worth remarking that no proof of Theorem $A$ avoiding the characteristic $p$ reduction is currently known.
This kind of argument was first used by Mori in order to prove the following theorem, which settles a conjecture due to Hartshorne:
Theorem B. If $X$ is a smooth complex projective variety of dimension $n$ with ample tangent bundle, then $X \cong \mathbb{P}^n.$
See [S. Mori, Projective manifolds with ample tangent bundle, Ann. of Math. 110 (1979)].
More details about Theorem $A$ (as well as its complete proof) can be found in the books [Debarre: Higher-dimensional algebraic geometry] and [Kollar-Mori, Birational geometry of algebraic varieties].
Best Answer
For question 2, consider the following scenario.
There are two mathematicians. Alice chooses a problem and comes up with $N$ possible approaches to solve it. Bob tries $N-1$ of the approaches and can't make them work, and reports this to Alice, who tries the $N$th approach, and succeeds. I think it's clear that for $N$ sufficiently large, Bob deserves coauthorship.
I am not sure exactly what the cutoff is, and it depends on unspecified details, but I think the large $N$ limit is fairly clear.