What is the probability of at least one ace and at least one heart for two draws without replacement? I have C(52,2) - C(48,2) - C(39,2) + C(36,2)/C(52,2). So, all the ways of drawing two cards from a deck of 52 minus all the ways of drawing an ace minus all the ways of drawing a heart plus all the ways of drawing neither an ace or heart, divided by all the ways of drawing two cards from a deck of 52. Is this correct? If not, why?
 A: This is correct and is a good example of how to use inclusion-exclusion properly, though your notation and typesetting could be improved.
$1 - \left(\dfrac{\binom{48}{2}+\binom{39}{2}-\binom{36}{2}}{\binom{52}{2}}\right)=\dfrac{29}{442}$
This is, of course, the same answer as what is suggested in the comments above... drawing the ace of hearts and any other card in some order, or drawing a nonheart ace and a nonace heart in some order:
$\dfrac{2}{52} + 2\cdot\dfrac{12}{52}\cdot\dfrac{3}{51}=\dfrac{29}{442}$
Be warned that when typing math with ascii, a phrase like a - b / c or even worse a - b/c like you had is more commonly interpreted as $a - \frac{b}{c}$ where the $a$ is not a part of the fraction which is clearly not what you intended.  Further, the notation $\binom{n}{k}$ for binomial coefficients is preferable to $C(n,k)$ to avoid confusion with other functions (at least in the west).
A: Although your attempt is correct after correcting typo, a direct approach  is simpler.
$\binom11\binom{51}1 = 51$ [ace of hearts and any other]
$\binom31\binom{12}1=36$ [non-heart ace and a non-ace hearts]
and $Pr = \Large\frac{87}{\binom{52}{2}}$

ADDED
It will be crystal clear if you look at the $4$ disjoint parts

*

*$12 \;\;{\heartsuit}$ non aces

*$1 \;\;\;\;{\heartsuit}$ ace

*$3 \;\;\;$ non ${\heartsuit}$ aces

*$36\;\;$ non ${\heartsuit}$ non aces

