Since $[x+3]=[x]+3$, we get $f(x)=[x{]}^2-[x]-6=([x]+2)([x]-3)$.

This is positive when $[x]\le -3$ or $[x]\ge 4$, i.e., $x\in (-\infty ,-2)\cup [4,\infty )$, so (1) is false.

It is negative when $[x]\in \{-1,0,1,2\}$, i.e., $x\in [-1,3)$, so (2) is correct.

${\int }_0^{2}f(x)dx={\int }_0^1(-6)dx+{\int }_1^2(-6)dx=-6-6=-12$, so (3) is false.

$f(x)=0$ when $[x]=3$ or $[x]=-2$, each giving an interval — infinitely many solutions, so (4) is false.