Problem 145

(1) $f(x)+f(2x+1)=|x-2|+|2x-1|= \begin{cases} 3-3x,x 2\\ \end{cases}$

When $x 2$, from $3x-3\geqslant 6$, we get $x\geqslant 3$.

Therefore, the solution set for the inequality $f(x)\geqslant 6$ is $(-\infty,-1]\cup[3,+\infty)$;

(2) Since $a+b=1(a,b > 0)$,

$\frac {4}{a}+ \frac {1}{b}=(a+b)( \frac {4}{a}+ \frac {1}{b})=5+ \frac {4b}{a}+ \frac {a}{b}\geqslant 5+2 \sqrt { \frac {4b}{a}\cdot \frac {a}{b}}=9$,

So for all $x\in R$, $f(x-m)-f(-x)\leqslant \frac{4}{a}+ \frac{1}{b}$ always holds if and only if: for all $x\in R$, $|x-2-m|-|-x-2|\leqslant 9$,

That is, $\max(|x-2-m|-|-x-2|)\leqslant 9$,

Since $|x-2-m|-|-x-2|\leqslant |(x-2-m)-(x+2)|=|-4-m|$,

So, $-9\leqslant m+4\leqslant 9$,

Therefore, $-13\leqslant m\leqslant 5$.

The final answer for the range of $m$ is $\boxed{-13\leqslant m\leqslant 5}$.