one end group with positve first Betti number$\beta^{(2)}_1(G)>0$ Could anyone give me an example of a countable finitely generated (f.g.) discrete group $G$ with one end but have non-trivial $H^1(G,\ell^2G)$? 
To be precise, consider the following two cases.
(1) Find a countable discrete f.g. non-amenable group G with $H^1(G,\mathbb{Z}G)=0$, but $\beta^{(2)}_1(G)\neq 0$.
(2) Find a countable discrete f.g. amenable group G with $H^1(G,\mathbb{Z}G)=0$, but $H^1(G,\ell^2(G))\neq 0$.
Any remarks, hint, references are welcomed!!
Thanks in advance!

Remarks:
1, It is well-known that a f.g. infinite group $G$ has 1-end iff $H^1(G, \mathbb{Z}G)=0$.
2, When $G$ is infinite, $H^1(G, \mathbb{Z}G)\hookrightarrow H^1(G,\ell^2(G))$.
3, When $G$ is a non-amenable countable discrete group,  $\beta^{(2)}_1(G)\neq 0\Longleftrightarrow H^1(G,\ell^2(G))\neq 0$ by corollary 2.4 in this paper. This is not true for amenable group, as can be seen for $G=\mathbb{Z}$.
4, I suspect for (1), we might take a torsion group with positive first Betti number, but I do not know why it has 1-end. Is it possible for a torsion free group to satisfies conditions in (1)?
5, For (2), it might be easy to find examples, but I do not know even for $G=\mathbb{Z}^2$, how to find $H^1(G,\ell^2(G))$.

(2) If $G$ is an {\em arbitrary} infinite amenable countable group, then $H^1(G,\ell^2(G))$ is nonzero (because it is not Hausdorff!: the subspace of coboundaries $B^1(G,\ell^2(G))$ is not closed in the space of cocycles $Z^1(G,\ell^2(G))$. Hence if $G$ is a non-virtually-cyclic amenable countable group, e.g. $\mathbf{Z}^2$, it answers (2).
(1) The surface groups $\Gamma_g$, $g\ge 2$, are examples.
