Expected number of cards you should turn before finding an ace

This is only an answer to your first question. I will show that your answer is correct by deriving the same answer in another way.

The probability that the seven of clubs turns up before the first ace is $\frac15.$ This is because each of the five relevant cards (the four aces and the seven of clubs) has the same one-in-five chance of being first.

Likewise, the probability that the jack of diamonds turns up before the first ace is $\frac15,$ and the same goes for the queen of hearts, the four of spades, and every other non-ace card in the deck. Since there are $48$ non-ace cards, the average number of (non-ace) cards preceding the first ace is $48\cdot\frac15.$

Imagine you have the following setup:

_A1_A2_A3_A4_

Each ace separated out evenly and we are interested in the pile that's before A1. For a standard deck of cards you have 52 cards - 4 aces = 48 cards left, and $$ \frac {48} 5 = 9.6$$ cards for each pile. So basically you would have to turn all 9.6 cards + the A1 card in order to see the first ace. So the answer is $$1 + \frac{48} {5} = 10.6 $$

Another way to do this with sums:

Suppose we draw all the cards in the deck. Let $k$ denote the place at which the first Ace appears. Note that $k$ can take integer values from $1 - 49$, because there must be at least $3$ other open spots for the other three Aces that will be drawn. We count the number of deals where the first Ace is in the $k^{th}$ spot as follows:

1. If we fix the first Ace is at the $k^{th}$ spot in the deal, then the other $3$ Aces must occur after the $k^{th}$ spot in the deal. This means there are $52 - k$ spots these Aces can occur, and thus $\binom{52 - k}{3}$ places orderings for this aces in the deal IGNORING suit.
2. To take in account suit, we multiply by $4!$ to account for all the permutations of the four suits of the aces.
3. Now that we accounted for the aces, we just need to consider all the other 48 cards left. For each ordering of the aces, there are $48!$ different ordering of the other cards in the remaining $48$ slots of the deal. We multiply again by $48!$ to account for this.
4. With that information, we know that since there are $52!$ total deals of the cards, meaning that the probability of getting the configuration with the first ace in the $k^{th}$ slot of the deal is: $\frac{\binom{52 - k}{3} \times 4! \times 48!}{52!}$
5. Since expectation is just the sum $\sum{n * Prob(n)}$ where $n$ is just the number of cards until we draw the first Ace (including the Ace), we can just take the sum $\sum_{n = 1}^{49}{n \times \frac{\binom{52 - k}{3} \times 4! \times 48!}{52!}} = \frac{1}{52*51*50*49}\sum_{n = 1}^{49}{\frac{n \times 4! \times (52 - n)!}{3! \times (49-n)!}} = \frac{1}{52*51*50*49}\sum_{n = 1}^{49}{4 \times n \times (52 - n)(51 - n)(50 - n)} = 10.6$