Can Euclidean TSP be exactly solved in time better than (sym)metric TSP?
One can exploit planar separator structures in a geometric setting and thus solve Euclidean TSP exactly in $O(n^{c \sqrt{n}})$ time, where $c$ is some small constant greater than 1. There's at least three simultaneous results for this; Smith's PhD thesis [1], Kann's PhD thesis [2], and Hwang et al. [3]. Each algorithm has a running time of $O^*(c^{\sqrt{n} \log n})$. This is much better than $O(n^2 2^n)$. As far as I know, it is still an open question whether the running time can be improved by say getting rid of the logarithmic term. 



[1] W.D. Smith, Studies in computational geometry motivated by mesh generation. Ph.D. Thesis, Princeton University, Princeton.

[2] V. Kann, On the approximability of NP-complete optimization problems, Ph.D. Thesis, Kungliga Tekniska Högskolan, Stockholm, 1992.

[3] R.Z. Hwang, R.C. Chang, R.C.T. Lee, The searching over separators strategy to solve some NP-hard problems in subexponential time,
Algorithmica 9 (1993) 398–423.