Find the number of rational solutions of the following equations (i.e., rational $x$ and $y$ satisfy the equations)
$$x^2+y^2=2$$$$x^2+y^2=3$$$\textbf{(A)}~2\text{ and }2$
$\textbf{(B)}~2\text{ and }0$
$\textbf{(C)}~2\text{ and infinitely many}$
$\textbf{(D)}~\text{Infinitely many and }0$
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 \(\boxed{\text{(B)}~2\text{ and }0}\)