2. Color the positive integers $1,2, \cdots, 15$ in blue or red. Find the number of coloring methods that satisfy the following conditions:
(1) The positive integer 15 is red;
(2) If the positive integers $x$ and $y$ are of different colors, and $x+y \leqslant 15$, then $x+y$ is colored blue;
(3) If the positive integers $x$ and $y$ are of different colors, and $x y \leqslant 15$, then $xy$ is colored red.
2. First, consider the case where 1 is red. Then from $2 \sim 15$ are all red.

Otherwise, assume some number $k$ is blue. By condition (3), we know $1 \cdot k=k$ is red, which is a contradiction. Thus, $1,2, \cdots, 15$ are all red, satisfying the conditions.
Now consider the case where 1 is blue.
$$
\begin{array}{l}
\text { By } 15=1+14=2+13=3+12 \\
=4+11=5+10=6+9=7+8,
\end{array}
$$

Then by condition (2), we know 1 and 14, 2 and 13, 3 and 12, 4 and 11, 5 and $10$, 6 and $9$, 7 and 8 are the same color.
(i) 2 is blue.
This is because if 2 is red, by condition (2), we know $3=1+2$ is blue, $5=3+2$ is blue, $\cdots \cdots 15=$ $13+2$ is blue, which contradicts condition (1). Thus, 2 is blue, and 13 is blue.
(ii) 7 is blue.
This is because if 7 is red, by condition (2), we know $8=1+7$ is blue, which contradicts the fact that 7 and 8 are the same color. Thus, 7 is blue, and 8 is blue.
(iii) 4 is blue.
This is because if 4 is red, by condition (2), we know $11=4+7$ is blue, which contradicts the fact that 4 and 11 are the same color. Thus, 4 is blue, and 11 is blue.
(iv) $3, 6, 9, 12$ are the same color.
This is because $12=3+9, 9=3+6$.
Thus, 15 is red, $1, 2, 4, 7, 8, 11, 13, 14$ are blue, 3, 6, 9, 12 are the same color, and 5, 10 are the same color.

Notice that, if 3 is red, then by condition (2), we know $5=2+3$ is blue, in which case, 3 and 5 cannot be the same color.

In summary, the total number of colorings that satisfy the conditions is 4, namely all red, only 15 is red, only numbers divisible by 5 are red, and only numbers divisible by 3 are red.