Fun with combinatorics and 80 business customers
There is one easier way to do it, which even lets you directly use the result from the first part as part of your calculations: "at least $2$" and "at most $1$" are complementary events, which means that the probability of one of them happening is equal to one minus the probability of the other happening, or written with more symbols: $$ P(\text{at least }2)=1-P(\text{at most }1) $$ And "at most $1$" takes a lot fewer steps to calculate (although technically, it's done in exactly the way you planned to calculate "at least $2$").
The denominators of the fractions stay constant. The total multiplication across the denominators is all the ways to pick 12 people from 80, where the order is retained: $$ 80\cdot 79\cdot 78\cdot 77\cdot 76\cdot 75\cdot 74\cdot 73\cdot 72\cdot 71\cdot 70\cdot 69 = \frac{80!}{68!} $$
We say that order is unimportant, so $\frac{80!}{68!\,12!} = {80 \choose 12}$ options
Then the numerators are the combination of the choices from the angry $(k)$ and non-angry $(12-k)$ groups, which are ${7 \choose k}$ and ${73 \choose 12-k}$, so overall the probability is $$\frac{{7 \choose k}{73 \choose 12-k}}{80 \choose 12}$$ and checking this against your result for $k=1$ we have $$\frac{{7 \choose 1}{73 \choose 11}}{80 \choose 12} = \frac{68!\,12!}{80!}\frac{7!}{1!\,6!}\frac{73!}{ 62!\,11!} = \frac{12\cdot 7\cdot 68\cdot 67\cdot 66\cdot 65\cdot 64\cdot 63}{80\cdot 79\cdot 78\cdot 77\cdot 76\cdot 75\cdot 74} \approx 0.413458415 $$
The easiest of the possible calculations is where none of the chosen employees are angry $(k=0)$, which is $$\frac{{7 \choose 0}{73 \choose 12}}{80 \choose 12} = \frac{68!\,12!}{80!}\cdot 1 \cdot\frac{73!}{ 61!\,12!} = \frac{68\cdot 67\cdot 66\cdot 65\cdot 64\cdot 63\cdot 62}{80\cdot 79\cdot 78\cdot 77\cdot 76\cdot 75\cdot 74} \approx 0.305171687$$
Then all other cases can be worked similarly.
You talk about getting a negative factorial when finding the combination of how to choose $2$ seats from $12$, but that is just ${12 \choose 2}= \frac {12!}{10!\,2!} = \frac{12\cdot 11}{2} = 66$ so I don't know how you arrived at a negative factorial.