13. A homemade deck of cards consists of 52 cards (including 4 suits: Hearts, Diamonds, Spades, and Clubs. Each suit has one card for each point value from 1 to 13). After shuffling, the cards are placed face down. To ensure that at least 2 cards have the same point value and color, you must draw at least $\qquad$ cards. If you need to ensure that among the drawn cards, there are at least 3 cards with consecutive point values (regardless of color), you must draw at least $\qquad$ cards.
13, $27 ; 37$

Solution: For the first case, one can take 1 card each of the red and black $1, 2, 3, 4, 5, 6, 7, 8, 9, 10$, 11, 12, 13 points, totaling $13 \times 2=26$ (cards). Taking one more card will definitely result in a pair with the same point value and color, thus ensuring that there are 2 cards with the same point value and color. This is the worst-case scenario, so at least 27 cards must be drawn to guarantee that 2 cards have the same point value and color.

For the second case, the following combinations are possible:
$$
(1,2,3), (4,5,6), (7,8,9), (10,11,12), 13 \text { . }
$$

Thus, for the 9 shaded numbers, taking all four suits, one can draw $(4 \times 2+1) \times 4=36$ (cards) without having 3 cards with consecutive point values.

Now consider drawing 37 cards. In the extreme case, 4 of these 37 cards are 13, leaving at least 33 cards to be drawn from the four groups (1,2,3), (4,5,6), (7,8,9), (10,11,12). According to the pigeonhole principle, at least 9 of these numbers must come from one of the groups, ensuring that there are 3 cards with consecutive point values in that group. Therefore, at least 37 cards must be drawn.