Are there infinite number of 3-braids with trivial closure? Not counting equivalent braids, are there finite or infinite numbers of 3-braids whose closures are trivial knot or links? If the answer is infinite, are there some patterns in those infinite numbers of braids, e.g. there exists some repeated parts?
 A: Suppose that $\beta$ is a three-braid whose closure is trivial.  Suppose that $\gamma$ is any three-braid.  Then $\gamma \beta \gamma^{-1}$ is again a three-braid with trivial closure.  However, if $\beta$ is non-trivial, then for "generic" $\gamma$, the braids $\beta$ and $\gamma \beta \gamma^{-1}$ will not be equivalent.
To explain the word generic, note that $\beta$ commutes with its powers (and its roots, if any).  Thus taking $\gamma$ to be a power (or a root, or a power of a root) of $\beta$ will result in $\gamma \beta \gamma^{-1}$ being equivalent to $\beta$.  More generally, $\gamma$ should not lie in the centraliser of $\beta$.  However, the centraliser is typically a very small subset of the braid group. (One notable exception to this is when $\beta$ lies in the centre of the braid group.)

Here is a simple version of HJRW's suggestion: Let $\beta$ be the braid $\sigma_1 \sigma_2^{-1}$.  This is a three-strand pseudo-Anosov braid.  Thus its centraliser (in the braid group) is virtually $\mathbb{Z}^2$ (as it contains $\beta$ and the central element $\Delta$).
A bit surprisingly, this is also the only pseudo-Anosov three-braid (conjugacy class) closing to the unknot, due to the work of Birman-Menasco (cited in the comments).
