so my solution was to...

a) find the number of hands where exactly one suit has exactly 3 cards
b) find the number of hands where exactly two suits have exactly 3 cards
c) add a) and b) then divide by c(52,8). 

so starting with b) because it's slightly easier...

c(4,1) way to pick a suit and c(13,3) ways to choose 3 cards in said suit...
c(3,1) way to pick rest of suits and c(13,3) ways to pick cards in that suit...
2 cards left among 26, so just c(26,2).

so answer for b) would be - > 12\*c(13,3)^2\*c(26,2)

then going with a)

c(4,1) way to pick a suit and c(13,3) ways to choose 3 cards in said suit...

c(39,5) ways to pick rest of cards but must consider that we want to exclude possibility of having another 3 in another deck...

so what i did was count the ways in which we could have 3 cards in a suit among the 3 suits and 5 cards we need...

so c(3,1) ways of choosing a suit along with c(13,3) ways of selecting 3, and c(26,2) ways of selecting rest of the suits...

so c(39,5) - 3\*c(13,3)\*c(26,2) is my answer for a). 

add them and divide by c(52,8) but i get a probability that is way too high...