Consider a deck of 52 cards (there are 13 cards of each suit). a. How many different ways can you a

Look at the first question. We are given 13 diamond cards. The 7,8,9,10 cards are supposed to stick together, either at the front or back of the line. Hence, we consider the possibilities separately:
1) The cards are at the front of the line. Then, these can be permuted amongst themselves in 4! ways, and the rest can be permuted amongst themselves in 9! ways so the the answer is 4!9!
2) The cards are at the back of the line. Here, it is important to see a symmetry, namely that if you take any arrangement of the first case and reverse it, you get an arrangement of the second case, and vice-versa. Hence, the answer here is 4!9! again.
So the answer would be 2*4!9!
We come to the second question.
First of all, the number of all possibilities when 8 cards are picked is $(\genfrac{}{}{0ex}{}{52}{8})$without observing the order of picking.
Next, we look at our favourable cases. We split them in two:
1) An ace and exactly three queens. The ace can be picked in four ways, and the three queens can be picked by excluding one queen, which is done in four ways again. We don't care what the rest are, these are picked in $(\genfrac{}{}{0ex}{}{48}{4})$ ways.
2) An ace and all four queens. The ace can be picked in four ways, and the four queens are fixed. We don't care what the rest are, these are picked in $(\genfrac{}{}{0ex}{}{48}{3})$ ways, because we have three remaining cards to pick out of 8.
Thus, the answer would be ${\displaystyle \frac{16\times <!--\; \times \; -->(\genfrac{}{}{0ex}{}{48}{4})+4\times <!--\; \times \; -->(\genfrac{}{}{0ex}{}{48}{3})}{(\genfrac{}{}{0ex}{}{52}{8})}}$
The strategies used in these questions is very simple:
1) Split the favourable case into exclusive events i.e. events that cannot occur simultaneously. For example, 7,8,9,10 cannot have been both the start and the end of the 13-card line, so we could examine them separately.
2) Know when we are permuting and when we are combining. The first question had to do with permutation, because in a line, the arrangement matters. However, In the second question, we do not care what order the aces,queens etc are picked in, so we have to take combinations.
3)The concept of symmetry is useful. It is nice to see when one situation can be transformed to the other symmetrically, as I did with the reversing in the first question. I should add that the second case could have been dealt with separately, but the answer would be the same.
4)Exclusive events are added, independent events are multiplied. I hope you get the difference.