If \( a = b \), then \( a + 3b \) is even. Therefore, \( a+3b=2 \), which implies that \( b<1\). Thus, in this case, there is no solution. Now, suppose without loss of generality that \( a > b \). If \( a \equiv 1 \pmod{3} \), then \( 2a + 1 \equiv 0 \pmod{3} \), which implies that \( 2a + 1 = 3 \), meaning \( a = 1 \), but this is not possible since \( a > b \). On the other hand, if \( a \equiv 2 \pmod{3} \), then \( a + 1 \equiv 3 \), which implies that \( a = 2, b = 1 \), \( a + 1 = 3 \), \( b + 1 = 2 \), \( 2a + 1 = 5 \), \( 2b + 1 = 3 \), and \( a + 3b = 5 \), \( 3a + b = 7 \). Finally, if \( a \equiv 0 \pmod{3} \), then \( a + 3b \) is not prime. We conclude that the only solutions are \( (a, b) = (2, 1) \) and \( (a, b) = (1, 2) \).