The optimal number of questions that could be achieved is 5 questions. I will demonstrate with only 6 questions (improving from 7 as stated on other answers).
As others have, determine suit in two questions:
- Is your color red?
If no to red, the color is black so next ask is your suit spades. If yes to red, ask if your suit is hearts
- Is your suit xxxx (per above) If yes, you know the suit from the question. If no, you know the suit from the only other suit with that color.
Thinking of the type of card as Ace, Two, Three, ..., ..., Queen, King:
- Are the number of letters in the type of card is less than, greater than or equal to 4?
If less than 4 letters, your card is an Ace, Two, Six, or Ten.
If more than 4 letters, your card is a Three, Seven, Eight or Queen.
If equal to 4 letters, your card is a Four, Five, Nine, Jack or King. I'll take each of these separately.
- Case less than 4 letters: Is your card value less than, greater than or equal to six? If equal or more, you know the card value, so proceed to ask for the card. If less, you have a 5th question for whether ace or two.
Now for the next case:
- Case more than 4 letters: Is your card value less than, greater than or equal to Eight? If equal or more, you know the card value, so proceed to ask for the card. If less, you have a 5th question for whether three or seven.
And the last case.
- Case equal to 4 letters: Is your card value less than, greater than or equal to Nine? If equal, you know the card value, so proceed to ask for the card. If less, you have a 5th question for whether four or five. If more, you have a 5th question for whether jack or king.
And finally the question for their card:
- (or this could be the 5th question): Combine the suit found above with the value determined to ask the exact card value.
Mathematically, the maximum values you can determine with a two value question (yes/no) is n=2^(q-1) where n is the number of items you are asking against and q is the number of questions asked. If the scenario was "how many questions until you know the answer, the formula would be n=2^q but the OP stated the last question had to be asking the value, so this takes one more question. In our case 1 question will be required if there is only one item queried against. 2 questions required for 2 items, 3 questions for up to 4 items, 4 questions for up to 8 items, (etc.).
However, that is not the case in our example. Since the OP allows for a 3 value answer (less than, equal or greater), the optimal equation is 3^(q-1). This means with optimal questions, you can split the number of possibilities by 1/3 for each question. 1q= 1 item, 2q= 3 items, 3q= 9 items, 4q = 27 items, 5q = 81 items. With optimal questions, 52 cards should be able to be found with just 5 questions. In this case I have demonstrated using 6 questions.