Assuming that during every shift all 12 people are involved in 4 groups of 3, it obviously takes at least 4 shifts before everyone will have worked at some point with every from all the other groups, because for everyone there are 8 such 'other' people, and during any shift they can only get to meet 2 people.
In fact, here is a schedule that does it in exactly 4 shifts:
\begin{array}{c|cccccccccccc}
Shift&a&b&c&d&e&f&g&h&i&j&k&l\\
\hline
1&1&2&3&4&1&2&3&4&1&2&3&4\\
2&1&2&3&4&2&1&4&3&3&4&1&2\\
3&1&2&3&4&3&4&1&2&4&3&2&1\\
4&1&2&3&4&4&3&2&1&2&1&4&3\\
\end{array}
(so, during every shift, group together the people with the same number)
For part $(a)$, you want to multiply $\binom{11}{4}$ and $\binom{10}{3}$, because for every choice of $4$ men, we could have any combination of $3$ women, and vice versa. This is the number of ways to get our $7$ team members. Now, we can assign the managers in $\binom{7}{3}$ ways, the accountants in $\binom{4}{2}$ ways, and the CFO $\binom{2}{1}$ ways. Once we've done this, the CEO is determined. So the answer is
$$\binom{11}{4}\binom{10}{3}\binom{7}{3}\binom{4}{2}\binom{2}{1}$$
For part $(b)$, observe that either one the CEO is one of those three (in which case he is the only one of those three on the team) or none of them are on the team. If the CEO is one of these three, we have $3$ possible choices for CEO, and $\binom{8}{3}$ ways to fill in the other male teammates. So there are $3\binom{8}{3}\binom{10}{3}$ teams of this form. We can assign the managers in $\binom{6}{3}$ ways and the accountants in $\binom{3}{2}$ ways.
If the CEO is anyone else, that means these three men are not on the team. In this case, we choose the males in $\binom{8}{4}$ ways and the females in $\binom{10}{3}$ ways, so there are $\binom{8}{4}\binom{10}{3}$ teams of this form. We still need to assign roles, so as in part $(a)$ we multiply by $\binom{7}{3}\binom{4}{2}\binom{2}{1}$. All in all, there are
$$3\binom{8}{3}\binom{10}{3}\binom{6}{3}\binom{3}{2} +\binom{8}{4}\binom{10}{3}\binom{7}{3}\binom{4}{2}\binom{2}{1}$$
such teams.
For part $(c)$, we take the total number of possible teams (which is $\binom{11}{4}\cdot\binom{10}{3}$ from part $(a)$), and subtract the number of teams where Jessica, Marnia and Ryan are all chosen. If these three are given, this means we can choose the remaining male team members in $\binom{10}{2}$ ways and the remaining female in $\binom{8}{1}$ ways. Accounting for role assignment, the number of such teams is
$$\left[\binom{11}{4}\binom{10}{3} -\binom{10}{2}\binom{8}{1}\right]\cdot\binom{7}{3}\binom{4}{2}\binom{2}{1}$$
Best Answer
This is a special case of what's commonly known as the Oddtown problem: a family of subsets of $\{1,2,\dots,n\}$ where every set has odd size, but every two sets intersect in an even number of elements, can have at most $n$ sets. Conversely, given a family of $n+1$ sets of odd size, two of them must have an odd intersection.
The complete solution to the Oddtown problem is given in an answer to the question I linked to above.
In the special case of this problem, every set (work team) has three elements (employees). If $n=100$ but there are $101$ sets in the family, two of them must have an odd intersection. We forbid work teams from sharing all three employees, so this must mean that two work teams share exactly one employee.
Here's a non-linear algebra solution. We assume that no two work teams have exactly one employee in common, and show that there's at most $100$ teams.
The key lemma is that "share $2$ members in common" must be transitive. If teams $A$ and $B$ share two employees, and so do teams $B$ and $C$, then both $A$ and $C$ contain two of $B$'s three members, so they overlap in at least $1$ employee: and therefore in two.
So we can split up the work teams into connected components, so that any two teams in each component share $2$ employees, and teams in different components share no employees. Now we show that if a component has $k$ teams, those teams together must include at least $k$ employees.
There are two kinds of components:
So now in each component there's at least as many employees as teams, and therefore overall there's at least as many employees as teams: for $101$ teams we'd need $101$ employees, which is a contradiction.