(1) To prove: $b_{n+1}-b_n=b_n=\frac{2}{2a_{n+1}-1}-\frac{2}{2a_n-1}=\frac{2}{2(1-\frac{1}{4a_n})-1}-\frac{2}{2a_n-1}=\frac{4a_n}{2a_n-1}-\frac{2}{2a_n-1}=2$

Therefore, the sequence $\{b_n\}$ is an arithmetic sequence, with $a_1=1$, $b_1=2$, hence $b_n=2n$. $\boxed{4\text{ points}}$

(2) $c_n=6^n+(-1)^{n-1}\lambda\cdot4^n$, for $c_{n+1}>c_n$ to always hold, then

$6^{n+1}+(-1)^n\lambda\cdot4^{n+1}>6^n+(-1)^{n-1}\lambda\cdot4^n\Rightarrow 6^n+(-1)^n\lambda\cdot4^n>0$

When $n$ is even, $\lambda>-\frac{6^n}{4^n}=-\left(\frac{3}{2}\right)^n$, thus $\lambda>\left[-\left(\frac{3}{2}\right)^n\right]_{\text{max}}=-\frac{9}{4}$

When $n$ is odd, $\lambda\leq\frac{6^n}{4^n}=\left(\frac{3}{2}\right)^n$, thus $\lambda<\left[\left(\frac{3}{2}\right)^n\right]_{\text{min}}=\frac{3}{2}$

Therefore, $\lambda\in\left(-\frac{9}{4}, \frac{3}{2}\right)$. $\boxed{9\text{ points}}$

(3) From (1), $\frac{1}{b_n(b_n+1)}=\frac{1}{2n(2n+1)}=\frac{4}{16n^2+8n}<\frac{4}{16n^2+8n-3}=\frac{4}{(4n-1)(4n+3)}=\frac{1}{4n-1}-\frac{1}{4n+3}$ (for $n\geq2$).

Therefore, $\frac{1}{b_1(b_1+1)}+\frac{1}{b_2(b_2+1)}+\ldots+\frac{1}{b_n(b_n+1)}<\frac{1}{6}+\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\ldots+\frac{1}{4n-1}-\frac{1}{4n+3}\right)=\frac{13}{42}-\frac{1}{4n+3}<\frac{13}{42}$. $\boxed{\text{End}}$