We make the assumption that the total number of people are "large" and so we may model our approach on selection with replacement rather than without.
Now... let us look at the probability that we know a very specific person.
You correctly found that this will be $.02\cdot .4 + .98\cdot .05=.057$
However, you have misinterpreted or misused this value as it appears in later calculations. You incorrectly used this as the probability that you know all three students at a table rather than how you should have used it which is the probability that you know a specific single student without regards to whether or not you know any others.
Now... the probability of knowing $i$ of the three people sitting with you at your table, we can model with a binomial distribution. To know a specific person occurs with probability $.057$ and to not know a specific person, this occurs with probability $.943$.
To know none of the people at your table, i.e. $i=0$, this occurs when you don't know the first person, you also don't know the second person, and we also don't know the third person. Thanks to our assumption on the number of people being "large" we can find the probability of all of these things happening simultaneously as being the product of these values, $(.943)^3$.
To know exactly one of the people at your table, i.e. $i=1$, this occurs when you don't know the first person or the second person but you do know the third, or you don't know the first and third person but do know the second, or you do know the first but not the second or third. Each of these values we find will be the same, so we can add the three of them together which can be written more simply as multiplying by three and will be $3\cdot (.943)^2(.057)^1$
More generally, we can use the binomial distribution to write this much more quickly and generally, letting $X$ be the random variable counting the number of people at the table we know, that $X=i$ as:
$$\Pr(X=i)=\binom{3}{i}(.057)^i(.943)^{3-i}$$
On to the next question, given that $i$ of the students at your table are dorm mates of yours, we are curious to find the probability that we know everyone at the table.
Well... with $i$ people that we are dorm mates with, we want to find the probability that we know each individually, and for the remaining $3-i$ people we want to find the probability that we know of them individually, and multiply the probabilities together.
$$(.4)^i\cdot (.05)^{3-i}$$
The third problem you should approach with Bayes' theorem and will wait to see if you understand what I've already written before continuing too far, but will involve you needing to calculate various other probabilities such as the probability that all three people are members of your dorm and you know all three, finding the probability that you know all three, and so on...
For the third: letting $A$ be the event that we know all three people and $B$ the event that all three people are dorm-mates of ours, we were tasked with calculating $Pr(B\mid A)$
$$Pr(B\mid A)=\dfrac{Pr(A\cap B)}{Pr(A)} = \dfrac{Pr(B)Pr(A\mid B)}{Pr(A)}=\dfrac{(.02)^3(.4)^3}{(.057)^3}$$
Best Answer
$P($Law student from B has at least one seat to her right$) = {2 \over 3}$
No Law students and no students from B can be to her right, so that discounts herself, the other 2 Law students and the other 2 students from B, leaving a choice of 4 students.
$P($Engineering student from A is in the seat immediately to the right$) = {1 \over 4}$
$P($Law student from B has exactly zero seats to her right$) = {1 \over 3}$
$P($Engineering student from A is in that non-existent seat$) = 0$ (just not possible!)
$P($Engineering student from A is immediately to the right of Law student from B$) = {2 \over 3} \times {1 \over 4} +{ 1 \over 3} \times 0 = {2 \over 12} = {1 \over 6}$