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Yes. There need not be a feasible region.
In both cases the constraints are used to produce an n-dimensional simplex which represents the "feasible region". In the case of linear programming this is the feasible region. But that is not the case for integer programming since only those points within the region for which the variables are integer are feasible.The objective function is then used to find the maximum or minimum - as required. In the case of a linear programming problem, the solution must lie on one of the vertices (or along one line in 2-d, plane in 3-d etc) of the simplex and so is easy to find. In the case of integer programming, the optimal solution so found may contain one or more variables that are not integer and so it is necessary to examine all the points in the immediate neighbourhood and evaluate the objective function at each of these points. This last requirement makes integer programming solutions more difficult to find.
Yes. If the feasible region has a [constraint] line that is parallel to the objective function.
Systems of inequalities in n variables with create an n-dimensional shape in n-dimensional space which is called the feasible region. Any point inside this region will be a solution to the system of inequalities; any point outside it will not. If all the inequalities are linear then the shape will be a convex polyhedron in n-space. If any are non-linear inequalities then the solution-space will be a complicated shape. As with a system of equations, with continuous variables, there need not be any solution but there can be one or infinitely many.
Yes, but only if the solution must be integral. There is a segment of a straight line joining the two optimal solutions. Since the two solutions are in the feasible region part of that line must lie inside the convex simplex. Therefore any solution on the straight line joining the two optimal solutions would also be an optimal solution.