The first part is a classical Stars and Bars problem.
For the second part substitute $x_6 = x_1 + 12$ and apply the Stars and Bars reasoning for $x_2 + x_3 + x_4 + x_5 + x_6 = 24$
For the third part do same as for part two, but this time substitute $x_6 = x_1 + 18$ and then subtract the number you will get from the number of total solutions, which we obtained in part one.
The final part is similar to the part three, but now you will need a little help from the Inclusion-Exclusion Principle
Let $S_{a, b, c, d, e} $ be the number of solutions with $x_1\ge a$, $x_2 \ge b$, $x_3\ge c$, $x_4\ge d$ and $x_5\ge e$.
The equation can be written as
$$(x_1-a)+(x_2-b)+(x_3-c)+(x_4-d)+(x_5-e)=21-a-b-c-d-e$$
So we have
$$S_{a, b, c, d, e} =\binom{21-a-b-c-d-e+4}{4}$$
if $a+b+c+d +e\le21$ and is $0$ otherwise.
The answer to this question is
$$S_{0, 1,15,0,0}-S_{4,1,15,0,0}-S_{0,4,15,0,0}+S_{4,4,15,0,0}=\binom{9}{4}-\binom{5}{4}-\binom{6}{4}+0=106$$
Best Answer
I like to think of such problems in terms of putting identical marbles into distinguishable boxes. Here there are $21$ marbles to be distributed amongst $5$ boxes, and $x_k$ is the number of marbles in box $k$. We first ensure that box $3$ gets at least $15$ marbles and box $2$ at least $1$ marble by simply putting $15$ marbles into box $3$ and $1$ into box $2$; this leaves $21-15-1=5$ marbles to be distributed amongst the $5$ boxes. If you like, you can think of this as replacing $x_2$ by $x_2'=x_2-1$ and $x_3$ by $x_3'=x_3-15$, and counting solutions to $$x_1+x_2'+x_3'+x_4+x_5=5\tag{1}$$ subject to the condition that all five unknowns be non-negative integers, with $x_1\le 3$ and $x_2'<3$.
Now use the standard stars-and-bars calculation to find the number of unrestricted solutions to $(1)$ in non-negative integers; it’s explained quite well in the linked article. Then use an inclusion-exclusion calculation to subtract the unwanted solutions, i.e., those with $x_1>3$ or $x_2\ge 3$. This will be fairly easy, since it’s impossible for $x_1$ and $x_2$ to be too large simultaneously.
In case you’ve not done anything like this before, counting the solutions with $x_1>3$ is simply counting those with at least $4$ marbles in box $1$ to begin with: you deal with this restriction exactly as I dealt with the original lower bounds on $x_2$ and $x_3$.