Problem 2:
8C2*9C3*5C1 = 11760

I think you calculated three scenarios separately and then added them to get your final answer. But actually here we have to find how many possible combination are there for someone to get 6 bars (2 orange, 3 red and 1 blue). So it will be multiplication not addition..

Problem 3: even I am getting 45 here..

We have 4 math teachers and 5 English teachers..
We have to choose a committee of 5 such that there are at least 3 math teachers on that committee..

it can be done in 2 diff ways..
1) 3 Math Teachers and 2 English Teachers = 4C3*5C2 = 4*10 = 40
2) 4 Math Teachers and 1 English Teacher = 4C4*5C1 = 1*5 = 5

Total 40 + 5 =45

Can someone please check where am i going wrong if 45 is not the correct answer?

I agree with Hardik on P2 & P3.

For P1, I too get 2860,
(12C2*40C1)+(12C3) = 2860

Agreed with above. For P1 I am getting as follows:

If three cards are chosen at random from a standard deck of playing cards, how many different ways are there to draw the three cards if at least two cards are a jack, queen or a king?

Total Jack, Queen, King in 52 cards = 12.

At least two cards are a jack, queen or a king means = Either 2 cards are from J, K, Q and 1 card is from remaining cards OR (+) all 3 cards are from J, Q, K.

Hence Total= 12C2 * 40C1 + 12C3 = 2860.