Number of arrangements of the $10\heartsuit|9\heartsuit|8\heartsuit|5\spadesuit|2\clubsuit$ Let's say that the original ordering of the cards is $10\heartsuit|9\heartsuit|8\heartsuit|5\spadesuit|2\clubsuit$ and then we shuffle the cards.
I declare these two random variables and I want to calculate the probability of their outcomes:
$X$-number of cards that stayed in the same place after the shuffle.
$Y$-number of the heart suit cards that stayed in the same place after the shuffle.
For $X$ I have looked at this as the probability of picking the right place
$$\small{Pr[X=0]=\frac{D(5)}{5!}, Pr[X=1]=\frac{5*D(4)}{5!}, Pr[X=2]=\frac{{5\choose{2}}*D(3)}{5!}, Pr[X=3]=\frac{{5\choose{3}}*D(2)}{5!},Pr[X=5]=\frac{1}{5!},Pr[X=4]=0}.$$
For $Y$ :
$$Pr[Y=0]=\frac{\small\text{number of derangements of the 3 hearts*}|\{\{5\spadesuit,2\clubsuit\},\{2\clubsuit,5\spadesuit\}\}|}{\text{number of permutations}}=\frac{D(3)\cdot2}{5!}$$
For $Pr[Y=1]$ I'm having trouble counting.
 A: To compute the probability distribution for $Y$:
Note:  throughout, $!n$ will denote the number of Derangements on $n$ letters.
$Y=0\quad $  We have three cases, according to how many of the black cards are fixed.  If $0$ black cards are fixed then there are $!5$ ways to do it. If exactly one black card is fixed then there are $!4$ for each of the two ways to choose the fixed black card.  And if both black cards are fixed then there are $!3$.  Thus $$P(Y=0)=\frac {!5+2\times !4+!3}{5!}=\frac 8{15}$$
$Y=1\quad $  There are $3$ ways to choose the fixed Heart.  For a fixed choice, we again have three cases.  If $0$ black cards are fixed then there are $!4$ ways to do it. If exactly one black card is fixed then there are $!3$ ways to do it for each choice of fixed black card, and if both black cards are fixed then there is only $1$ way to do it.  Thus $$P(Y=1)=3\times \frac {!4+2\times 3!+1}{5!}=\frac 7{20}$$
$Y=2\quad$  There are $3$ ways to choose the fixed pair of Hearts.  For a fixed choice, we again have three cases.  If $0$ black cards are fixed then there are $!3$ ways to do it.  If exactly one black card is fixed then there are $!2$ ways to do it for each choice of fixed black card.  If both black cards are fixed then there is no way to do it.  Thus $$P(Y=2)=3\times \frac {!3+2\times !2}{5!}=\frac 1{10}$$
$Y=3\quad$  In this case we either permute the two black cards or we fix them both, so $$P(Y=3)=\frac 2{5!}=\frac 1{60}$$
Sanity checks:  first of all, these must sum to $1$.  Indeed we get $$\frac {32+21+6+1}{60}=1$$
Secondly, it is easy to see that the expected number of fixed hearts is $\frac 35$ (hint: use indicator variables).  And it is easy to confirm that number directly from the probabilities.
