What is Fermi surface and why is this concept so useful in metals research? What is Fermi surface and why is this concept so useful in metals research?
Particularly, I can somewhat appreciate the Fermi energy idea - the radius of Fermi surface which is a sphere. But is there any quantitative use of more complicated Fermi surfaces? 
 A: I think what you might be interested in are Van Hove singularities or the critical points of the Fermi surface, where the density of states as given by $dN/dK_{|k=k_f|}$ diverges. Now $dN/dk$ is proportional to the inverse of the gradient of the energy $ dN/dk \propto 1/\nabla E $. The locations with the greatest d.o.s. on the Fermi surface will exhibit singularities in various absorption and emission spectra. These are precisely those locations where the Fermi surface fails to be a smooth, differentiable surface.
Critical surfaces can be 0D (Fermi point), 1D (line), 2D (Fermi surface). Complicated substances (such as the High $T_c$ superconductors for e.g. which are composed of layers of cuprates) will in general have complicated Fermi surfaces. Because critical surfaces are topological entities they are robust with respect to small perturbations of the microscopic Hamiltonian of the system. In other words the critical surfaces determine the universality class to which the given surface belongs. The universality class determines whether a given material is a superconductor, ferromagnet, Mott insulator etc. Clearly to take a material from one universality class to another requires that one cross a phase boundary. Such a change also requires that the Fermi surface topology undergo a change. Consequently changes in Fermi surface topology can be understood as signs of phase transitions.
A Fermi surface is also important for reasons other than its topology, which describes the global characteristics of the material. Others will likely mention some of the complementary local aspects of the Fermi surface.
A: One more thing about geometrical properties of the Fermi surface. Its structure defines material transport properties like conductivity. It is actually equal to the integral of the mean free path along those wave vectors that define a Fermi surface.
Knowing this is very important. How do you sample the Fermi surface of a given metal?  By means of the de Haas-van Alphen effect.
A: The Fermi surface is simply the surface in momentum space where, in the limit of zero interactions, all fermion states with (crystal) momentum $|k|<|k_F|$ are occupied, and all higher momentum states are empty. Amazingly, Luttinger and Ward showed that the Fermi surface survives even with interactions to all orders in perturbation theory (Oshikawa later showed this nonperturbatively, see also the arXiv version).
The point of the Fermi surface is that this is where all of the low-lying excitations of the system live -- the Fermi energy is so much larger than room temperature that for room-temperature experiments, all of the thermodynamics is dominated by excitations right at the Fermi surface, and thus knowing its structure is very important.
A more advanced reason for the importance of the Fermi surface is that by knowing its structure (look up "Fermi surface nesting"), we can understand the instabilities of a metal, for example, for sufficiently low temperature, a normal metal settling into a charge density wave state.
A: A Fermi surface is one of constant energy $E_F$ (named the Fermi energy) surface in the momentum k-space where the energy below the Fermi energy $E_F$ is filled with occupied states. The momentum k-space can be viewed as an effective way to organize the Hilbert space of weakly interacting fermions. 
