Problem 1211

1. **Analyzing the first equation $x^2 + y^2 = 2$:**
- We need to find rational solutions $(x, y)$ such that $x^2 + y^2 = 2$.
- Notice that $x^2$ and $y^2$ must be non-negative rational numbers.
- Let's consider the possible values for $x^2$ and $y^2$. Since $x^2$ and $y^2$ are rational, they can be written as $\frac{a}{b}$ and $\frac{c}{d}$ respectively, where $a, b, c, d$ are integers and $b, d \neq 0$.
- For simplicity, let's check if there are integer solutions first. If $x$ and $y$ are integers, then $x^2$ and $y^2$ must be integers.
- The only pairs of integers $(x, y)$ that satisfy $x^2 + y^2 = 2$ are $(\pm 1, \pm 1)$.
- Therefore, there are 4 possible pairs: $(1, 1), (1, -1), (-1, 1), (-1, -1)$.
- Since these pairs are rational, we have 4 rational solutions for the first equation.

2. **Analyzing the second equation $x^2 + y^2 = 3$:**
- We need to find rational solutions $(x, y)$ such that $x^2 + y^2 = 3$.
- Consider the equation modulo 4. Note that the squares of any integer modulo 4 are 0 or 1.
- Therefore, $x^2 \equiv 0 \pmod{4}$ or $x^2 \equiv 1 \pmod{4}$ and similarly for $y^2$.
- The possible sums of two squares modulo 4 are $0 + 0 = 0$, $0 + 1 = 1$, $1 + 0 = 1$, and $1 + 1 = 2$.
- Notice that 3 is not among these possible sums. Hence, there are no integer solutions to $x^2 + y^2 = 3$.
- Since rational solutions would imply that $x^2$ and $y^2$ are rational numbers whose sum is 3, and since we have shown that there are no integer solutions, there cannot be any rational solutions either.

Conclusion:
- The number of rational solutions for $x^2 + y^2 = 2$ is 4.
- The number of rational solutions for $x^2 + y^2 = 3$ is 0.

The final answer is $\boxed{\text{(B)}~2\text{ and }0}$