Is martingale problem interesting?
Lets start with the basics. Suppose that $X(t)$ is a time-homogeneous Markov process. We define the transition semigroup of this process by $S(t)f(x) = \mathbb{E}[f(X(t)) \,|\, X(0) = x]$. (i.e. we have $S(0)$ is the identity and $S(t) S(r) = S(t+r)$ by the Markov property).
We can calculate the infinitesimal generator of this semigroup, which is the derivative of the semigroup at $t=0$: $Lf = \frac{d}{dt}|_{t = 0} S(t)f = \lim_{t \rightarrow 0} \frac{S(t)f - f}{t}$ defined for $f \in \mathcal{D}(L)$, the set of functions for which this limit exists. See http://en.wikipedia.org/wiki/C0-semigroup.
Skipping technicalities, we get by the vector valued version of the fundamental theorem of calculus that for $f \in \mathcal{D}(L)$ $$ S(t)f - f - \int_0^t L S(r)f d r = 0, $$ which implies that for $f \in \mathcal{D}(L)$
$$ f(X(t)) - f(X(0)) - \int_0^t L f(X(s)) d s $$ is a martingale.
$L$ is an important object because $L$ tells you what the process is doing. For example we have a process $X(t)$ on $\mathbb{R}$ such that $$ L^Xf(x) = f''(x) + b f'(x) $$ then the process is a diffusion process with drift $b$. A process $Y(t)$ on $\mathbb{N}$ such that its generator satisfies $$ L^Y f(n) = f(n+1) - f(n) $$ is a Poisson jump process.
In other words, every(reasonably behaved) time-homogeneous Markov process gives you a generator $L$.
For applications, one might want go the other way around. Suppose you want to model a phenomenon that you expect to behave Markovian. A first step is to write down an operator $L$ that should serve as the generator of the Markov process. However, a priori it is not clear that every operator $L$ corresponds to a Markov process. This is the Martingale problem: given an operator $(L, \mathcal{D}(L))$, find a process $X(t)$ such that for $f \in \mathcal{D}(L)$ $$ f(X(t)) - f(X(0)) - \int_0^t L f(X(s)) d s $$ is a martingale. If you succeed in doing this, then you have a process that behaves as your operator $(L,\mathcal{D}(L))$ describes.
A strategy to solve the martingale problem is by taking approximating generators $L_n$, i.e. $L_n \rightarrow L$ for some topology on the space of operators, for which it is known that there are Markov process $X^n(t)$ solving the martingale problem for $L_n$(for example the $L_n$ are bounded). Associated to these Markov processes $X^n(t)$ there are probability measures on the trajectory space, lets denote them with $\mathbb{P}^n$. Now one has to do two things. First, one shows that the sequence $\mathbb{P}^n$ has limit points[existence]. Second, one shows that there is at most one limit point[uniqueness].
The fact that $L_n \rightarrow L$ and some technicalities then implies that this unique limit point is a solution to the martingale problem for $L$.
Ps. One can also try to show that $(L,\mathcal{D}(L))$ generates a semigroup via the Hille-Yosida theorem, and then construct a Markov process from the semigroup. This is a different and approach, but gives essentially the same result. Depending on the operator one approach is easier than the other.
First of all, martingale problems often arise in the theory of Markov processes as they are related to the characterization. For example, let $(X_n)_{n\in \Bbb N}$ be a sequence of Markov processes with generators $\mathscr L_n$ and we would like to check whether they converge to a Markov process $X$ with a generator $\mathscr L$. In such a case, one can just check a certain convergence $\mathscr L_n\to\mathscr L$. However, to extend this result from $\mathscr L$ to $X$ one needs to show that a martingale problem for $\mathscr L$ has a unique solution. In general, one can define Markov processes as solutions of martingale problems.
Martingales themselves are not only (or almost not at all) about martingale problems (which themselves are more about operators). They often appear in optimal control problems thanks to their almost-constant structure, and are inevitable when dealing with SDEs or SPDEs. The latter often arise in evolutional dynamics and can have examples in all systems that evolve in time.
P.S. your reminder of what is a martingale problem, especially a linear operator applied to a process is misleading.