In quantum annealing, is it necessary that we need to end up in the ground state of the system hamiltonian? Is it necessary that we need to end up in the ground state of the system hamiltonian in order get the solution. What happens if we do not end up in the ground state? Can I still get the solution by some means?
 A: This depends on what you want to know, but generally no, you don't need to get exactly the ground state.
Quantum annealing can, to a large degree, be thought of as a way of finding the minimum of a polynomial.
For example, we might want to find the values of $x_i$ to minimize the polynomial
$$\sum_i h_i x_i + \sum_{ij} J_{ij}x_i x_j$$
where all $x_i \in \{-1, 1\}$.
A quantum annealer is a physical system whose Hamiltonian matches this polynomial (the values $x_i$ are the states of the qubits), let the system go to its ground state, and then read the values $x_i$ directly from the system.
Suppose the annealer gets close to the ground state, but not exactly to it.
For example, perhaps all the $x_i$ match the true ground state except for one of them.
We can ask, is this a problem?
To answer that question, consider the point of the minimization problem in the first place.
Suppose, for example, that we're trying to find a solution to the travelling salesman problem where the goal is to find the shortest route connecting a set of cities.
In this case, the $x_i$ encode the best answer, i.e. the shortest path.
If we get only one of the $x_i$ wrong, then our solution is still quite efficient!
Since the point is to find the route that takes the least time (or the least fuel, etc.), then it's ok in practice if we don't get exactly the best possible route.
This same reasoning can be applied to most minimization problems.
To summarize, it's usually fine to get a solution that's close to the true ground state because the point of annealing is to solve a minimization problem, and in practice it's totally fine if we don't get the absolutely best possible solution in minimization problems.
