Here is an approach to understanding your problem in greater generality.

Suppose we would like to count the number of *non-negative* and *ordered* integer solutions of the Diophantine equation $x_1 + x_2 + x_3 + x_4 + x_5 = 36$ with the following additional constraints: $x_1 \geq 4$, $x_3 = 11$ and $x_4 \geq 7$. First, as my colleague Mike suggests, let's subtract $x_3$ from both sides. We are then to count: $x_1 + x_2 + x_4 + x_5 = 25$ given said constraints.

I prefer to think about this problem combinatorially in terms of lattice points of dilated polytopes. The unconstrained inequality $x_1 + x_2 + x_4 + x_5 \leq R$ represents the number of non-negative lattice points of the $R$-dilate of the polytope $\mathbf{conv}(\mathbf{0}, \mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3, \mathbf{e}_4)$, which is bounded by a $4$-simplex or pentachoron and the natural boundaries of the positive orthant, where $\mathbf{e}_i$ represents the $i^{\text{th}}$ basis of $\mathbb{R}^{4}$. For simplicity, we'll just call this object a *simplicial $4$-polytope* $P$.

It is clear that the number non-negative solutions of $x_1 + x_2 + x_4 + x_5 = R$ is found by considering the inequality $R - 1 < x_1 + x_2 + x_4 + x_5 \leq R$ which can be determined by simply subtracting the number of lattice points of two consecutive dilates of the same pentachoron. That said, the number of non-negative lattice points of an $R$-dilate of $P$ is the iterated sum,
\begin{align}
N(R) = \sum_{i_1 = 0}^{R} \sum_{i_2 = 0}^{R - i_1} \sum_{i_3 = 0}^{R - i_1 -i_2} \sum_{i_4 = 0}^{R - i_1 -i_2 -i_3} 1 = \binom{R + 4}{R}.
\end{align}
So, with no constraints, we have the following number of non-negative solutions:
\begin{align}
\binom{R+4}{R} - \binom{R + 3}{R - 1}.
\end{align}
To account for the constraints, we modify the various limits of summation accordingly,
\begin{align}
N(R) = \sum_{i_1 = 4}^{R} \sum_{i_2 = 0}^{R - i_1} \sum_{i_3 = 7}^{R - i_1 -i_2} \sum_{i_4 = 0}^{ R-i_1 - i_2 - i_3} 1.
\end{align}
So the answer to your first question (without a constraint on $x_5$) is therefore $N(25) - N(24)$, which is $680$. To account for the constraint on $x_5$, we further modify the innermost summation,
\begin{align}
N(R) = \sum_{i_1 = 4}^{R} \sum_{i_2 = 0}^{R - i_1} \sum_{i_3 = 7}^{R - i_1 -i_2} \sum_{i_4 = 0}^{ \min(5, R-i_1 - i_2 - i_3)} 1.
\end{align}
The answer to your second question is therefore $N(25) - N(24)$, which is $515$.

To compute the solutions explicitly, use the following mathematica code:

FindInstance[x1 + x2 + x4 + x5 == 25 && x1 >= 4 && x2 >= 0 && x4 >= 7 && x5 >= 0, {x1, x2, x4, x5}, Integers, 1000]}

FindInstance[x1 + x2 + x4 + x5 == 25 && x1 >= 4 && x2 >= 0 && x4 >= 7 && 5 >= x5 >= 0, {x1, x2, x4, x5}, Integers, 1000]}

Compute the "Length[]" of the output and you can verify that there are indeed $680$ and $515$ integral solutions, respectively.