These problems can be solved by using ordered or unordered selections. Some of your answers are incorrect since you used ordered selections without taking all possible favorable ordered selections into account.
As stated in the question, we have a $32$ card deck consisting of the eight ranks 7, 8, 9, 10, J, Q, K, A in each of the four suits clubs, diamonds, hearts, spades.
What is the probability of selecting a heart and a spade when two cards are drawn from the deck?
Method 1: Using unordered selections.
Since there are $32$ cards in the deck and we are drawing without replacement, there are $\binom{32}{2}$ possible hands. For the favorable cases, we must select one of the eight hearts and one of the eight spades. Hence, the probability of selecting a heart and a spade is
$$\Pr(\text{a heart and a spade}) = \frac{\dbinom{8}{1}\dbinom{8}{1}}{\dbinom{32}{2}}$$
Method 2: Using ordered selections.
In your calculation, you found the probability of selecting a heart, then a spade. However, it is possible to select a spade, then a heart. Therefore, you should have obtained
$$\Pr(\text{a heart and a spade}) = \Pr(\color{red}{\heartsuit})\Pr(\spadesuit \mid \color{red}{\heartsuit}) + \Pr(\spadesuit)\Pr(\color{red}{\heartsuit} \mid \spadesuit) = \frac{8}{32} \cdot \frac{8}{31} + \frac{8}{32} \cdot \frac{8}{31}$$
What is the probability of selecting two hearts when two cards are drawn from the deck?
Your answer is correct. The answer can also be obtained using combinations. We must select two of the eight hearts while selecting two of the $32$ cards in the deck. Hence,
$$\Pr(\text{two hearts}) = \frac{\dbinom{8}{2}}{\dbinom{32}{2}}$$
What is the probability of selecting two red cards when two cards are drawn from the deck?
Your answer is correct. The answer can also be obtained using combinations. We must select two of the $16$ red cards while selecting two of the $32$ cards in the deck. Hence,
$$\Pr(\text{two red cards}) = \frac{\dbinom{16}{2}}{\dbinom{32}{2}}$$
What is the probability of selecting exactly one spade when two cards are drawn from the deck?
Method 1: Using unordered selections.
We must select one of the eight spades and one of the $24$ non-spades while drawing two of the $32$ cards in the deck. Hence,
$$\Pr(\text{exactly one spade}) = \frac{\dbinom{8}{1}\dbinom{24}{1}}{\dbinom{32}{2}}$$
Method 2: Using ordered selections.
You calculated the probability of selecting a spade, then a non-spade. However, you could also select a non-spade and then a spade.
$$\Pr(\text{exactly one spade}) = \Pr(\spadesuit)\Pr(\spadesuit^C \mid \spadesuit) + \Pr(\spadesuit^C)\Pr(\spadesuit \mid \spadesuit^C) = \frac{8}{32} \cdot \frac{24}{31} + \frac{24}{32} \cdot \frac{8}{31}$$
What is the probability of selecting a heart or an ace when two cards are drawn from the deck?
Method 1: We subtract the probability of selecting two cards that are not hearts or aces from $1$.
A heart or an ace is selected unless both selected cards are neither aces nor hearts. The deck contains eight hearts and four aces, including the ace of hearts ($\color{red}{A\heartsuit}$), which would be counted twice if we simply added the number of aces to the number of hearts. Therefore, there are $8 + 4 - 1 = 11$ cards in the deck that are hearts or aces. The remaining $21$ cards in the deck are neither hearts nor aces. Hence,
$$\Pr(\text{a heart or an ace}) = 1 - \frac{\dbinom{21}{2}}{\dbinom{32}{2}}$$
Method 2: We correct your approach.
There are $8 + 4 - 1 = 11$ cards in the deck that are hearts or aces. Therefore, the probability of selecting a heart or ace on the first draw is $11/32$. If the first draw is not a heart or an ace, which occurs with probability $21/32$, the probability that the second card is a heart or an ace is $11/31$. Hence,
$$\Pr(\text{a heart or an ace}) = \frac{11}{32} + \frac{21}{32} \cdot \frac{11}{31}$$
