possibility of choosing three from a pool and choosing three from a different pool At the local zoo, a new exhibit consisting of 3 different species of birds and 3 different species of reptiles is to be formed from a pool of 8 bird species and 6 reptile species. How many
exhibits are possible if
a. there are no additional restrictions on which species can be selected?
b. 2 particular bird species cannot be placed together (e.g., they have a predator-prey
relationship)?
c. 1 particular bird species and 1 particular reptile species cannot be placed together?
(A) Because we're choosing three from 8 and three from 6, I got 
8*7*6 + 6*5*4. Is my reasoning correct?
(B) For reptiles we get the same as above (6*5*4) but for birds, because I got
8*6*5 because two birds cannot be paired with together. The complete answer would be 8*6*5 + 6*5*4
(C) 8*7*6 + 5*4*3 because reptile numbers depend on what we choose for birds.
Is my reasoning for above answers correct?

It does not matter which order the species are selected. Thus we will (usually) need to divide by $3!$ for the count of ways of selecting the bird and reptile species, something which you didn't do for all the questions.
You also added the counts for the birds and reptiles, which is wrong (it would be true if you were choosing either the birds or the reptiles, instead of both at once). Instead you should multiply them. For (a) you should get an answer of $\frac{8×7×6}{3×2×1}×\frac{6×5×4}{3×2×1}$; this forms the base count, which we will rely on in solving (b) and (c).
The other two questions are better solved by counting the number of possibilities that are not allowed, then subtracting from the base count. For (b), if we select the forbidden pair of birds, we have $6$ possibilities for the remaining bird, so there are $6×\frac{6×5×4}{3×2×1}$ disallowed selections.
For (c), if we select the forbidden reptile/bird pair, we still have $2$ birds to select from $7$ and $2$ reptiles from $5$. Again, order does not matter, so we divide by $2$ for each. The number of disallowed selections is $\frac{7×6}2×\frac{5×4}2$.
