5. Among the following four propositions:
(1) The sum of a rational number and an irrational number is always an irrational number;
(2) The sum of two positive irrational numbers is always a positive irrational number;
(3) There is always an irrational number between any two irrational numbers;
(4) There do not exist irrational numbers \(a, b\) such that \(a^b\) is a rational number. The number of true propositions is ( ).
(A) 1
(B) 2
(C)?
(I) 4
5.B.
(1) and (3) are true, (2) and (4) are false.

If $a$ is a rational number, $\alpha$ is an irrational number, and $a+\alpha=b$ is a rational number, then $a=b-a$ is a rational number, which is a contradiction;
The sum of the positive irrational number $\sqrt{2}$ and $2-\sqrt{2}$ is a rational number;
Let $\alpha, \beta$ be irrational numbers, $\alpha<\beta$. It is easy to see that $\alpha<\frac{\alpha+\beta}{2}<\beta$. If $\frac{\alpha+\beta}{2}$ is irrational, the proposition holds. If $\frac{\alpha+\beta}{2}$ is rational, by (1), $\alpha+\frac{\alpha+\beta}{2}$ is irrational, and $\alpha<\frac{1}{2} \cdot\left(\alpha+\frac{\alpha+\beta}{2}\right)<\frac{\alpha+\beta}{2}<\beta$, so $-\frac{1}{2}\left(\alpha+\frac{\alpha+\beta}{2}\right)$ is the desired number;

If $(\sqrt{3})^{\sqrt{2}}$ is a rational number, the proposition is already negated. If $(\sqrt{3})^{\sqrt{2}}$ is irrational, then $\left(\sqrt{3}^{\sqrt{2}}\right)^{\sqrt{2}}=(\sqrt{3})^{2}=3$ is a rational number. Taking $a=(\sqrt{3})^{\sqrt{2}}, b=\sqrt{2}$; the proposition is also negated.