Constrained Maximization problem involving integer variables The maximization problem may be seen as a problem of integer programming problem and may be solved by Gomory cut method or branch and bound technique.
Is there any other way to see this problem?

 A: Combine and simplify your constraints. We see for instance from the second and last that
$$(x_2 + x_4) + x_5 > (x_1 + x_3) > (x_2+x_4) - x_5.$$
We can conclude that $x_5 \geq 2$, since $x_5=1$ is not possible (it would force $x_2+x_4=x_1+x_3 = \frac{499}{2})$. From combining the first two constraints we also have $x_1+x_3+x_5 > 500$.
Let's ignore the other constraints for now and find an upper bound on the solution by considering only the equality constraint
$$(x_1+x_3) + (x_2+x_4) + x_5 = 1000,$$
as well as $(x_1+x_3) + x_5 > 500$ and $x_5 \geq 2$. Observe that it's optimal to set $x_5=2$; increasing $x_5$ decreases the base $(x_1+x_3)$ of the objective function at no benefit to the exponent $(x_2+x_4)$. We then have that $x_1+x_3 \geq 499.$
You can show with some elementary calculus that $x^{k-x}$ on $x\in [499,\infty]$ is maximized at $x=499$ whenever $k$ is sufficiently small $(k \leq 499+499\log 499 \approx 3600)$.
We now have an upper bound on the solution: $499^{499}$, with $x_1+x_3=499$, $x_2+x_4=499$, $x_5=2$. The base cannot be smaller than $499$ and it's not optimal to set it larger at the cost of the exponent (by the above argument).
Now it's a matter of checking that this upper bound can be achieved within the region carved out by the constraints. Only the middle three inequalities are still relevant and they reduce to
\begin{align*}
x_1 &> 1\\
1+x_2 &> x_1\\
x_2 &< 498
\end{align*}
which (along with $x_1+x_3=499$ and $x_2+x_4=499$) are clearly feasible; and from these you can address part 2 of your problem.
A: Substitute $x_5 = 1000- (x_1 + \cdots + x_4)$ into the constraints to get
\begin{eqnarray}
500 &>& x_2 + x_4 \\
&\vdots& \\
500 &>& x_1 + x_3 \\
\end{eqnarray}
It is easy to check that if the constraints are satisfied then $x_5 > 0$.
This gives an upper bound of $(x_1+x_3)^{x_2+x_4} \le 499^{499}$, and since this is attained with $x_1+x_3= 499, x_2+x_4 = 499, x_5=2$ this gives the answer to (a) using $x_1=2, x_2=497, x_3=497,x_4=2, x_5 = 2$.
For (b) start by substituting $x_5 = 2$ into the constraints gives (note $x_k \in \mathbb{N}$ as well)
\begin{eqnarray}
499 &=& x_2 + x_4 \\
497 &\ge& x_3 \\
499 &\ge& x_4+x_1 \\
497 &\ge& x_2 \\
499 &=& x_1 + x_3 \\
\end{eqnarray}
The above also show that $x_1 \ge 2, x_4 \ge 2$.
If we set $x_1+x_4 = 4$ we get one solution.
If we set $x_1+x_4 = 5$ we get 2 solutions.
If we set $x_1+x_4 = 499$ we get 496 solutions.
Hence the total number of solutions is $1 + \cdots + 496 = \binom{497}{2}$.
