Rajkumar's direct solution works nicely. You mentioned that the problem could be considered as conditional probability, and I'll show here that that method gives the same answer.
As I mentioned in a comment, the main reason for the inaccuracy of your original answer is overcounting the number of ordered $6$-card hands with the first four being non-aces, and at least one of the last two being aces.
Let $A$ be the event that the player is dealt $6$ non-ace cards that are randomly drawn from a standard deck of $52$ cards.
Let $B$ be the event that the first four of the player's $6$ cards are not aces.
$P(B \mid A$) is clearly $1$, since the first four must be non-aces if the six are all non-aces.
Since each card is chosen independently without replacement, we can find the probability of never getting an ace as
$$P(A) = \frac{48 \cdot 47 \cdot 46 \cdot 45 \cdot 44 \cdot 43}{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48 \cdot 47}$$
Likewise
$$P(B) = \frac{48 \cdot 47 \cdot 46 \cdot 45}{52 \cdot 51 \cdot 50 \cdot 49}$$
Now we can use Bayes' rule:
$$P(A\mid B) = \frac{P(B \mid A) \cdot P(A)}{P(B)}$$
But $P(B \mid A)$ is $1$, so
$$P(A\mid B) = \frac{P(A)}{P(B)}$$
$$P(A\mid B) = \frac{44 \cdot 43}{48 \cdot 47}$$
To justify your answer for Question 1: Let's count the number of ways the Ace of Diamonds could be in the deck. There would have to be $6$ other cards in the deck, so there are $\binom{11}{6}$ ways this can happen (and the remaining $5$ cards must be in your opponent's hand. There are $\binom{12}{7}$ total ways the cards could be distributed between the deck and the opponent's hand if we don't care where the Ace of Diamonds is. So the probability the Ace of Diamonds is in the deck is:
$$\frac{\binom{11}{6}}{\binom{12}{7}}=\frac{\frac{11!}{6!5!}}{\frac{12!}{7!5!}}=\frac{11!7!}{12!6!}= \frac{7}{12}.$$
As others have stated, your formula for Question 2 is correct, but you made an arithmetic error.
For Question 3, let $B=\{\text{other player's hand has at most one spade}\}$ and $A=\{\text{Ace of Diamonds is in the deck}\}$. As you said, you need to compute $P(B\mid A)$. Since the event $B$ is somewhat complicated - the other player's hand could have zero or one spade - but $B^c$ is simple, because it means the other player's hand has both spades. Now you can use the fact that $P(B\mid A) = 1-P(B^c \mid A)$. Now you just need to compute $P(B^c \mid A)=P(B^c \cap A)/P(A)$. The numerator involves counting the number of ways that both spades are in the opponent's hand and the Ace of Diamonds is in the deck, which I believe you should be able to do.
Best Answer
It is not correct to say that whether you are told that an ace is in the hand or whether you are told that the ace of spade is in the hand, the probability of having at least two aces is the same
Case 1: You are told that you have at least one ace
Note that in the following formula, the sample space shrinks to $\binom{52}5 - \binom{48}5$ to take into account that at least one ace must be present, and that in the numerator we need to ensure that one ace is there
$Pr = 1 - \dfrac{\binom41\binom{48}4}{\binom{52}5 - \binom{48}5},\;\approx 0.1222$
Case 2: You are told that you have the ace of spades
Sinca ace of spades is assured, the sample space is now simply $\binom{51}{4}$, and from the numerator, we just subtract $\binom{48}4$ to indicate that more aces must be present.
Thus $Pr = 1 - \dfrac{\binom{48}4}{\binom{51}4}\; \approx 0.2214$
Added approach
Since probability problems are often "tricky", here is an approach simpler to understand, though longer computationally.
Basically, the numerator will count hands with $2$ or more aces, and the denominator, hands with $1$ or more aces
Case 1: You are told that you have at least one ace
$Pr = \dfrac{\binom42\binom{48}3 + \binom43\binom{48}2 +\binom44\binom{48}1}{\binom41\binom{48}4+\binom42\binom{48}3 + \binom43\binom{48}2 +\binom44\binom{48}1}\approx 0.1222$
Case 2: You are told that you have the ace of spades
Bear in mind that you have the ace of spades and are left to pick from $51$ cards including $3$ aces
$Pr = \dfrac{\binom31\binom{48}3 + \binom32\binom{48}2 +\binom33\binom{48}1}{\binom30\binom{48}4+\binom31\binom{48}3 + \binom32\binom{48}2 +\binom33\binom{48}1} \approx 0.2214$