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As hinted by user2661923, we can answer the first 2 questions as follows:

Let's say the 4 persons are $A_1, A_2, A_3$ and $A_4$, meaning that the order of drawing cards is $A_1, A_2, A_3$ and then $A_4$.

Also let's assume that we draw the cards 8 times instead of 4.

And we count only the first 4 draws.

Let's number the draws from 1 to 8.

For each draw (e.g. draw 3), each card has equal chance to be in that draw.

Therefore the unique card has a probability of $\frac{1}{8}$ to be taken in each draw.

Q1: What is the probability that one of the players will draw the unique card?

Answer: This happens when the unique card is among the first 4 cards to be drawn.

The answer is $4 \times \frac{1}{8}= \frac{1}{2}$

Q2: What is the probability that the fourth player will draw the unique card?

Answer: This happens when the unique card is in draw 4.

The answer is $\frac{1}{8}$

As hinted by user2661923, we can answer the first 2 questions as follows:

Let's assume that we draw the cards 8 times instead of 4.

And we count only the first 4 draws.

Let's number the draws from 1 to 8.

For each draw (e.g. draw 3), each card has equal chance to be in that draw.

Therefore the unique card has a probability of $\frac{1}{8}$ to be taken in each draw.

Q1: What is the probability that one of the players will draw the unique card?

Answer: This happens when the unique card is among the first 4 cards to be drawn.

The answer is $4 \times \frac{1}{8}= \frac{1}{2}$

Q2: What is the probability that the fourth player will draw the unique card?

Answer: This happens when the unique card is in draw 4.

The answer is $\frac{1}{8}$

As hinted by user2661923, we can answer the first 2 questions as follows:

Let's say the 4 persons are $A_1, A_2, A_3$ and $A_4$, meaning that the order of drawing cards is $A_1, A_2, A_3$ and then $A_4$.

Also let's assume that we draw the cards 8 times instead of 4.

And we count only the first 4 draws.

Let's number the draws from 1 to 8.

For each draw (e.g. draw 3), each card has equal chance to be in that draw.

Therefore the unique card has a probability of $\frac{1}{8}$ to be taken in each draw.

Q1: What is the probability that one of the players will draw the unique card?

Answer: This happens when the unique card is among the first 4 cards to be drawn.

The answer is $4 \times \frac{1}{8}= \frac{1}{2}$

Q2: What is the probability that the fourth player will draw the unique card?

Answer: $\frac{1}{8}$, because the unique card has probability $\frac{1}{8}$ to be in draw 4.

As hinted by user2661923, we can answer the first 2 questions as follows:

Let's say the 4 persons are $A_1, A_2, A_3$ and $A_4$, meaning that the order of drawing cards is $A_1, A_2, A_3$ and then $A_4$.

Also let's assume that we draw the cards 8 times instead of 4.

And we count only the first 4 draws.

Let's number the draws from 1 to 8.

For each draw (e.g. draw 3), each card has equal chance to be in that draw.

Therefore the unique card has a probability of $\frac{1}{8}$ to be taken in each draw.

Q1: What is the probability that one of the players will draw the unique card?

Answer: This happens when the unique card is among the first 4 cards to be drawn.

The answer is $4 \times \frac{1}{8}= \frac{1}{2}$

Q2: What is the probability that the fourth player will draw the unique card?

Answer: This happens when the unique card is in draw 4.

The answer is $\frac{1}{8}$

As hinted by use2661923, we can answer the first 2 questions as follows:

Let's say the 4 persons are $A_1, A_2, A_3$ and $A_4$, meaning that the order of drawing cards is $A_1, A_2, A_3$ and then $A_4$.

Also let's assume that we draw the cards 8 times instead of 4.

And we count only the first 4 draws.

Let's number the draws from 1 to 8.

For each draw (e.g. draw 3), each card has equal chance to be in that draw.

Therefore the unique card has a probability of $\frac{1}{8}$ to be taken in each draw.

Q1: What is the probability that one of the players will draw the unique card?

Answer: This happens when the unique card is among the first 4 cards to be drawn.

The answer is $4 \times \frac{1}{8}= \frac{1}{2}$

Q2: What is the probability that the fourth player will draw the unique card?

Answer: $\frac{1}{8}$ when the unique card is the 4th card to be drawn.

As hinted by user2661923, we can answer the first 2 questions as follows:

Let's say the 4 persons are $A_1, A_2, A_3$ and $A_4$, meaning that the order of drawing cards is $A_1, A_2, A_3$ and then $A_4$.

Also let's assume that we draw the cards 8 times instead of 4.

And we count only the first 4 draws.

Let's number the draws from 1 to 8.

For each draw (e.g. draw 3), each card has equal chance to be in that draw.

Therefore the unique card has a probability of $\frac{1}{8}$ to be taken in each draw.

Q1: What is the probability that one of the players will draw the unique card?

Answer: This happens when the unique card is among the first 4 cards to be drawn.

The answer is $4 \times \frac{1}{8}= \frac{1}{2}$

Q2: What is the probability that the fourth player will draw the unique card?

Answer: $\frac{1}{8}$, because the unique card has probability $\frac{1}{8}$ to be in draw 4.