1. Given that $a, b, c$ are the three side lengths of a triangle. The following conclusions are given:
(1) Three segments with lengths $a^{2}, b^{2}, c^{2}$ can form a triangle;
(2) Three segments with lengths $\sqrt{a}, \sqrt{b}, \sqrt{c}$ can form a triangle;
(3) Three segments with lengths $|a-b|, |b-c|, |c-a|$ can form a triangle;
(4) Three segments with lengths $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ can form a triangle;
(5) Three segments with lengths $k a, l b, m c (k+l=m, k, l, m \in \mathbf{N}_{+})$ can form a triangle. Among them, the incorrect conclusion numbers are
$=1$.(1)(3)(4)(5).
When $a=3, b=4, c=5$, $a^{2}+b^{2}=c^{2}$, then the lengths $a^{2}, b^{2}, c^{2}$ cannot form a triangle. Therefore, conclusion (1) is incorrect.

Assume $a>b>c>0$, and satisfies $a^2=b^2+c^2$, then $\sqrt{b}+\sqrt{c}>\sqrt{a}$.
Similarly, $\sqrt{c}+\sqrt{a}>\sqrt{b}, \sqrt{a}+\sqrt{b}>\sqrt{c}$.
Hence, the lengths $\sqrt{a}, \sqrt{b}, \sqrt{c}$ can form a triangle. Therefore, conclusion (2) is correct.

When $a=2, b=3, c=4$, $|a-b|+|b-c|=|c-a|$, then the lengths $|a-b|, |b-c|, |c-a|$ cannot form a triangle. Therefore, conclusion (3) is incorrect.

When $a=2, b=4, c=5$, $\frac{1}{a}>\frac{1}{b}+\frac{1}{c}$, then the lengths $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ cannot form a triangle. Therefore, conclusion (4) is incorrect.

When $a=3, b=4, c=5, k=3, l=4, m=7$, $k a+l b=9+16<35=m c$. Then the lengths $k a, l b, m c$ cannot form a triangle. Therefore, conclusion (5) is incorrect.