The formulas being used are not quite correct. Note that the problem (at least as first and currently stated) says that the $x_i$ are bigger than $0$.
a) Give $5$ marbles to Kid $1$. We have $13$ marbles left, to be distributed between $4$ kids, at least one to each. By the usual Stars and Bars argument, we need to choose $3$ gaps from the $12$ available to put a separator into. There are $\binom{12}{3}$ ways to do this. Another kind of analysis will yield the equivalent answer $\binom{12}{9}$.
You have $\binom{15}{12}$, which is not equal to $\binom{12}{3}$.
(b) You used the right strategy, count all ways, subtract the ways in which $x_4=5$. Let's count these excluded cases. So we need to distribute $13$ marbles between $3$ kids, at least $1$ to each kid. There are $\binom{12}{2}=\binom{12}{10}$ ways to do this. You have something else. The first term is also not right, it should be $\binom{18}{3}$ or $\binom{18}{15}$. The issue here, as usual, is forgetting about the condition $x_i\gt 0$.
c) Now we have $11$ marbles, to be distributed between Kid $1$ and Kid $4$. Hardly need formulas for this. Since each kid has to get at least $1$ marble, there are $10$ ways to do this. You got $12$, which would be the right answer if we allow the possibility of a kid getting no marbles. However, that is ruled out by the beginning of the problem, which says $x_i\gt 0$.
d) There are $4$ ways to distribute $5$ marbles between Kids $1$ and $2$, at least $1$ to each kid. That leaves $13$ marbles to be distributed between Kids $3$ and $4$, at least one to each kid. There are $12$ ways to do this, for a total of $(4)(12)$. You got $(6)(14)$, which would be correct if we allow a kid to have no marbles. But that is excluded.