(b) You want the number of solutions of $$x_1+x_2+x_3+x_4+x_5=20\tag{1}$$ in non-negative integers satisfying $x_1,x_2\ge 2$ and $x_3,x_4,x_5\ge 3$. Let
$$\begin{align*}
&y_1=x_1-2,\\
&y_2=x_2-2,\\
&y_3=x_3-3,\\
&y_4=x_4-3,\text{ and}\\
&y_5=x_5-3\;.
\end{align*}$$
Then $\langle x_1,\dots,x_5\rangle$ is a solution to $(1)$ meeting the extra conditions if and only if $\langle y_1,\dots,y_5\rangle$ is a solution in non-negative integers to
$$y_1+y_2+y_3+y_4+y_5=7\;.\tag{2}$$
If you’ve seen the interpretation of the original problem as counting the number of ways to put $20$ indistinguishable balls into $5$ distinguishable boxes subject to the condition that the first two boxes each contain at least $2$ balls and the others at least $3$, the new problem is what you get if you put the minimum requirement into each box and then ask how many ways there are to distribute the remaining $7$ balls without any restrictions.
Now you can use the formula that you know for the unrestricted case to get a total of $$\binom{7+5-1}{5-1}=\binom{11}4$$ solutions.
You can use the same basic technique to solve (a) or any other such problem when the restrictions are all lower bounds.
For (c) and other problems with upper bounds you’ll want an inclusion-exclusion argument. In this problem it’s not at all bad. First count the number of unrestricted solutions, and then subtract the number of ‘bad’ solutions, i.e., the number that have at $x_3\ge 12$. Note that counting these ‘bad’ solutions is a problem with lower bounds, so you now know how to handle it. The problem would be a bit messy if it were possible to violate more than one upper bound, but it’s not, since you have only one upper bound to worry about.