The one-sentence explanation is that cohomology on a space $X$ answers the question of when you can promote local solutions of a problem to global solutions; that is, if you can solve a problem on each element of a cover of $X$, can you patch those individual solutions together to get a global solution?
For de Rham cohomology, the first question is one of closed versus exact differential forms. Given a closed form $\omega$ on a smooth and (let's say) compact manifold $X$, is there another form $\eta$ with $d\eta = \omega$? The answer is easy on a star-shaped patch of $X$: just integrate. (I'm glossing over reduced versus unreduced cohomology, so you have to be more careful with the dimension of $\omega$ to get a definitive answer here.) With a bit more work, the same answer holds for a coordinate patch diffeomorphic to $\mathbb{R}^n$. That means we know the answer locally on $X$, but it's not clear how to proceed to a global answer. The classical example is $\mathbb{R}^2\setminus{0}$, where the form
\begin{align*}
d\theta = \frac{- y\, dx + x\, dy}{x^2 + y^2}
\end{align*}
is closed but not exact by Stokes' theorem, since its integral over a loop around $0$ doesn't vanish. That last part is a hint that topology is involved, since we can look at $\int_{\gamma} d\theta$ as a function of $\gamma$ and consider how it varies under smooth homotopy.
Still, the idea of closed versus exact differential forms may seem a bit arbitrary. One way of resolving it is to note that de Rham cohomology group $H^0(X)$ is the space of functions that are locally constant (as there are no (nonzero, I guess) forms of negative degree). Under some reasonable assumptions on $X$, that means that $H^0(X)$ is just the $\mathbb{R}$-space of connected components of $X$. The higher groups have similar topological ramifications that would take a bit too long to describe, but the point is that despite the de Rham cohomology defined in terms of a certain set of differential equations, they're fundamentally topological objects.
To that end, the answer to the question of what the de Rham cohomology groups are good for might be that they're a particularly useful and concrete way of defining and computing the cohomology of a space. There are multiple cohomology theories (not all of which are well-defined for all spaces): singular cohomology, cellular cohomology, etc. Because of abstract nonsense, these construction generally coincide under reasonable conditions on $X$. The short explanation is that all of these start with the same $H^0$ and satisfy certain axioms that force the higher $H^*$ to agree.
Quicker and simpler explanations for the specific questions you asked:
1) Some assorted applications:
De Rham cohomology is invariant under homotopy equivalence (in the smooth category), which allows us to easily prove that two manifolds are not homeomorphic or homotopy equivalent. Furthermore, cohomology comes with a ring structure that gives an extra condition to require for equivalence.
Homotopy groups are hard to compute; cohomology groups are generally much easiser. The Hurewicz theorem states that the lowest homotopy and homology group of a reasonable space coincide if they're in dimension $\geq 2$; in dimension $1$, you have to abelianize $\pi_1$ first. (Homology groups can generally be obtained from cohomology via, say, Poincare duality.)
Rational homology groups (i.e., $\pi_* \otimes \mathbb{Q})$ can be obtained from $H_*(X) \otimes \mathbb{Q}$ via minimal models, under some mild restrictions on the underlying space.
Closed, simply-connected $4$-manifolds are determined by their intersection forms, which in the smooth case is similar to the the map $H^2(X) \otimes H^2(X) \to \mathbb{R}$ defined by $(\alpha, \beta) \to \int_X \alpha \wedge \beta$. (It's only "similar to" in part because de Rham cohomology is naturally defined over $\mathbb{R}$, while we need to work over $\mathbb{Z}$ here.) With a lot of work, this leads to proving that there exist closed, simply-connected manifolds that have no smooth structure.
For an Eilenberg-MacLane space $K(G, A)$ with $A$ abelian, the map $f \to f^* \omega$ for some specific $\omega$ gives a bijection between the space $[X, K(G, n)]$ of maps $X \to K(G, n)$ modulo homotopy and $H^*(X, G)$. This leads to the idea of classifying spaces for vector bundles, and bundles over a given space can be classified via cohomology. This result also has deeper consequences; for example, it leads to a proof (although there are other ones available) that every closed, oriented $3$-manifold is the boundary of some compact, oriented $4$-manifold.
For groups $G$ and $A$ with $A$ abelian, a central extension of $G$ by $A$ is a group $E$ for which $A\subset Z(E)$ and $E/A = G$. These groups are classified by the second group-homology group $H^2(G, A)$. The connection with group-cohomology and topological-cohomology is that, for reasons of abstract nonsense, $H^*(G)$ is precisely $H^*(K(G, 1))$, with $K(G, 1)$ the Eilenberg-MacLane space above. (We're now definitely dealing with, e.g., singular topology rather than de Rham cohomology, since we have no convenient smooth structure at hand, but the point is that an abstract question about group theory turns into a problem about topology.)
In general, there are varieties of cohomology in commutative algebra, number thoery, physics, etc. At some point, it's like the concept of integral: the original impetus and most concrete application may be computing the area under a curve, but there are so many applications of it in so many different fields that it's hard to nail down one single thing that you can point and say it's designed for. Similarly, just as Riemann integrals are a first, concrete step toward a general idea of integration (leading to integrating differential forms, the Lebesgue integral, etc.), de Rham cohomology is one particular useful and concrete kind of cohomology theory.
2) Five is a lot of examples to give, and I'm not sure what exactly you're looking for here. There are a bunch of examples in Hatcher (in the topological rather than smooth case, but the principle and the axioms are effectively the same), and you can work through the known case of the genus $g$ oriented, closed surface.
