12. Given that $a, b$ are distinct positive numbers, insert two sets of numbers: $x_{1}, x_{2}, \cdots, x_{n}$ and $y_{1}, y_{2}, \cdots, y_{n}$, such that $a, x_{1}, x_{2}, \cdots, x_{n}, b$ form an arithmetic sequence, and $a, y_{1}, y_{2}, \cdots, y_{n}, b$ form a geometric sequence, then among the following inequalities:
(1) $\frac{1}{n} \sum_{k=1}^{n} x_{k}>\sqrt{a b}+\left(\frac{\sqrt{a}-\sqrt{b}}{2}\right)^{2}$,
(2) $\frac{1}{n} \sum_{k=1}^{n} x_{k}>\frac{a+b}{2}$,
$\sqrt[n]{y_{1} y_{2} \cdots y_{n}}<\sqrt{a b}$, (4) $\sqrt[n]{y_{1} y_{2} \cdots y_{n}}<\frac{a+b}{2}-\left(\frac{\sqrt{a}-\sqrt{b}}{2}\right)^{2}$, the ones that hold are $\qquad$
12. It is easy to know that $\frac{1}{n} \sum_{k=1}^{n} x_{n}=\frac{a+b}{2}=\sqrt{a b}+\frac{(\sqrt{a}-\sqrt{b})^{2}}{2}>\sqrt{a b}+\left(\frac{\sqrt{a}-\sqrt{b}}{2}\right)^{2}$, hence (1) holds, (2) does not hold.

It is also easy to know that $\sqrt[n]{y_{1} y_{2} \cdots y_{n}}=\sqrt{a b}=\left(\frac{\sqrt{a}+\sqrt{b}}{2}\right)^{2}-\left(\frac{\sqrt{a}-\sqrt{b}}{2}\right)^{2}<\frac{a+b}{2}-\left(\frac{\sqrt{a}-\sqrt{b}}{2}\right)^{2}$, hence (4) holds, (3) does not hold. Therefore, the answers are (1)(4).