8 pokers hands are dealt from a shuffled deck without replacement.
a. Find the probability that at least one of the 8 hands is a heart flush(all five cards are hearts).
Pr(at least one of 8 hands is heart flush) $= 1 -$ Pr(none of eight hands is heart flush) $=$
$$ \left( 8\frac{\binom{13}{5}}{\binom{52}{5}}-28\frac{\binom{13}{5}}{\binom{52}{5}}\frac{\binom{8}{5}}{\binom{47}{5}}\right)$$
Is this answer correct?
b. Find the expected value and variance of the total number of eight hands which are heart flushes.
Expected value $= 8\dfrac{13 \choose 5}{52 \choose 5}$
Variance $= 8\dfrac{13 \choose 5}{52 \choose 5}\left(1- \dfrac{13 \choose 5}{52 \choose 5}\right)$
Are these answers correct?
Let $A$ be the probability the first player has a heart flush. Then $A=\frac{13 \choose 5}{52 \choose 5}.$
Let $B$ be the probability the first and the second player both have heart flushes. Then $B=\frac{13 \choose 10}{52 \choose 10}.$
The probability at least one player has a heart flush is then ${8 \choose 1}A-{8 \choose 2}B=8A-28B$ which is in effect what you have written for question (a).
The probability of zero heart flushes is $1-8A+28B$, of one $8A-56B$ and of two $28B$.
Your expected value of $8A$ in (b) is correct. You can see this directly, or as $0\times(1-8A+28B) +1\times(8A-56B)+2\times 28B$.
Your variance of $8A(1-A)$ or $8A(1-8A)$ is slightly wrong. You could work out the variance to be $0^2\times(1-8A+28B) +1^2\times(8A-56B)+2^2\times 28B - (8A)^2 = 8A(1-8A)+56B$.