(Topological) Superconductor is an Insulator 
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*Superconductor may be viewed as an Insulator, if the Superconductor is formed by the BCS pairing where the excitations around the Fermi surface are gaped everywhere. However, superconductor also superconducts. So how can a superconductor be both an Insulator, as well as a superconductor?

*Topological Superconductor may be viewed as an Insulator, if the Superconductor is formed by the BCS pairing where the excitations around the Fermi surface are gaped everywhere. However, Topological superconductor also superconducts. So how can a Topological superconductor be both an Insulator, as well as a superconductor?

*The classifications of Topological Superconductor requires the assumption that the phases of matter to be gapped in the energy spectrum between ground state and 1st excited states, such as the work of Kitaev and Ryu et all. If Topological superconductor also superconducts, it should be gapless, should the classifications fail when Topological Superconductor becomes gapless and superconduct?
 A: 1. Answer to questions 1 and 2:
The energy gap that is most commonly referred to in a superconductor is the gap in the dispersion of Bogoliubov quasiparticles. These quasiparticles do not contribute to superconductivity. The supercurrent, which is composed of Cooper pairs, is responsible for superconductivity. For any non-zero temperature below the critical temperature ($T_{\rm c}$), the conduction electrons in a metal (which now superconducts) contribute to the formation of two types of quasiparticles:


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*Cooper pairs (bound state of two electrons), or

*Bogoliubov quasiparticles (superposition of electron and hole)


It is the Cooper pairs that flow without any resistance. Despite the existence of a non-zero density of (non-superconducting) Bogoliubov quasiparticles at finite temperatures, the reason the resistance remains zero below $T_{\rm c}$ is that the Cooper pair condensate "shorts out" the Bogoliubov quasiparticles. In other words, in terms of a circuit diagram, you have a finite resistor (Bogoliubov quasiparticles) in parallel with a zero resistor (Cooper pair condensate).
Ironically, the energy gap helps in maintaining superconductivity. Here's why: below $T_{\rm c}$, when we break a Cooper pair, by supplying an energy equal to the energy gap, we get a pair of Bogoliubov quasiparticles (not electrons). If the gap were zero (everywhere in the Brillouin zone) we would not have Cooper pairs, thus no supercurrent, and thus no superconductivity!
2. Answer to question 3:
No, it is still possible to classify gapless or nodal superconductors in terms of the Altland-Zirnbauer symmetry classes. Note that these systems only have gap vanishing in certain regions (points or lines) in the Brillouin zone; these regions are called nodes.
The classification of nodal systems, however, differs from the original work of Kitaev, Ryu et al. from ~2008-2010, which does indeed limit itself to fully gapped systems. Nonetheless, Ryu, Schnyder et al. later on extended this classification to nodal systems in 2013. This work can be found in the following paper:

S. Matsuura, P. Y. Chang, A. P. Schnyder, S. Ryu. "Protected boundary states in gapless topological phases." New Journal of Physics. 2013 Jun 3;15(6):065001. (arXiv)

If you're interested, you can find the aforementioned work on classification of nodal systems recently (Aug 2016) contextualized in the review article:

C. K. Chiu, J. C. Teo, A. P. Schnyder, S. Ryu. "Classification of topological quantum matter with symmetries." Reviews of Modern Physics. 2016 Aug 31;88(3):035005. (arXiv)

