I will answer your second question because it's the one with which I'm more familiar.
The question we're answering is: "Why does current in a superconductor move with no resistance?"
To understand this we should first understand why normal metals have nonzero resistivity. Imagine an electron in the metal and suppose it is traveling in some direction. If the electron never interacted with anything else then it would just go along merrily in that direction and in fact the resistance would be zero. However, in a normal metal the traveling electrons interact ("collide") with the ions of the metal via the Coulomb force because they're both charged. This transfers energy and momentum from the electron to the ion. Since that ion is tightly connected to the other metal ions, this energy and momentum transfer causes the lattice of ions to vibrate. An excitation of this kind of vibration of the lattice is called a "phonon". When you will hear people say that, in a normal metal, electrons scatter off of phonons, this is what they're talking about.
By the way, excitation of phonons is precisely heating of the metal, so we see that as the electron loses energy the metal heats up. This is Joule heating.
So, we can say that in a normal metal the resistivity is non-zero because the electrons scatter off of phonons. In a superconductor you'd think the situation would be the same. There are still vibrational modes in the metal, and the electrons and ions are still charged so of course they still interact via the Coulomb force. To understand why this doesn't happen we have to more carefully consider what's going on in a normal metal.
When an electron scatters off of a phonon the electron's state (ie. it's direction and speed of travel) changes. Electrons are in a class of particles called Fermions. Fermions are subject to the Pauli exclusion principle, which is that it is impossible to have two Fermions occupying the same state. That means, for example, that you can't have two electrons in the same momentum state. That means that in order for an electron to scatter off of a phonon, we have to have a situation where there is an unoccupied final state into which the electron can scatter. Otherwise, the scattering process simply cannot happen. In a normal metal at finite temperature, the electrons occupy a set of momentum states up to a certain level. Think of it as follows. When you put the first electron in the system it goes in the lowest energy state (which is at zero momentum). Because of the exclusion principle the next electron goes in a higher energy state, which also has more momentum. As you continue to add electrons you fill of a 3D ball of momentum states. This is called the "Fermi Sea" and is illustrated in part a of the attached diagram.
At zero temperature the electrons would fill the ball up to a cutoff energy. This is shown in dark red in the diagram. However, at finite temperature there's a bit of thermal energy that allows the electrons to jump around to slightly higher energy (and momentum) states. This little band is shown in pink in the diagram. In part b we zoom in on the pink band. The dark circles indicate filled electron states and the open circles indicate empty electron states. When an electron interacts with a phonon, it can only scatter and change state if there is a final state available. The green arrow indicates a possible scattering process, while the gray arrow indicates one forbidden by the exclusion principle.
In a superconductor, something really interesting happens. Because of interactions with the phonons, it turns out that there's an effective attractive interaction between the electrons! Basically what happens is that a traveling electron with its negative charge causes the positively charged ions of the metal to get sucked together a little bit. The electron leaves really fast, but the ions take longer to move around, so after the electron is gone there is still a region of increased density of positively charged ions. This causes other electrons to get attracted to that point, because of the higher positive charge. In this way, there is a weird time dependent attraction between electrons.
The attraction between electrons makes it turn out that there's a lower energy state than the filled red region shown in part a of the figure. This lower energy state is such that the usual single electron excitations are at an energy level that is a significant distance away from this new ground state. The gap in energy, shown in part c of the figure, is the reason that superconductivity happens. Now if electrons were to scatter off of a phonon, the only available states they have to scatter into are a large distance away in energy. That means that you need really high energy phonons (or something else) to disturb this ground state. So, as long as you don't whack your superconductor too hard, there's is NO scattering, and thus no resistance.
This argument also explains why superconductors have zero resistance at DC, but nonzero resistance at AC. If you put in high frequency perturbations, you can introduce enough energy to actually kick a superconducting electron out of the gap into those available states in the pink region. Remember, the energy of a perturbation is related to frequency by $E=h f$.
Summary: Superconducting metals have zero resistance because there aren't any available states for electrons to scatter into. No scattering means no resistance.
A superconductor conducts electricity without resistance because the supercurrent is a collective motion of all the Cooper pairs present.
In a regular metal the electrons more or less move independenly. Each electron carries a current $-e \textbf{v}(\textbf{k})$, where $\textbf{k}$ is its momentum and $\textbf{v}(\textbf{k}) = \partial E(\textbf{k})/\partial \textbf{k}$ is the semiclassical velocity. If an electron gets scattered from momentum $\textbf{k}$ to $\textbf{k}'$ it gives a corresponding change in the current. A sequence of such processes can cause the current to degrade.
In a superconductor, the story is totally different because the Cooper pairs are bosons and are condensed. This means that Cooper pairs self-organize into a non-trivial collective state, which can be characterized by an order parameter $\langle \Psi(\textbf{x}) \rangle = \sqrt{n} e^{i\theta(\textbf{x})}$ (where $\Psi$ is the annihilation operator for Cooper pairs.) which varies smoothly in space. Since the current operator can be written in terms of $\Psi$ it follows that gradients of $\theta$ give rise to currents of the condensate: $\textbf{j} = n(\nabla \theta + \textbf{A})$. All the small-scale physics (such as scattering) gets absorbed into the effective macroscopic dynamics of this order parameter (Landau-Ginzburg theory).
One should think of every single Cooper pair in the system taking part in some kind of delicate quantum dance, with the net effect being a current flow. But this dance is a collective effect and so it's not sensitive to adding or removing a few Cooper pairs. Therefore, scattering processes don't affect the current.
Best Answer
Both of them are super-interesting to study:-)
Superfluid is a low-temperature state of a quantum many-body system of electrically neutral particles (e.g. atoms). Superfluids have some amazing properties. For example, there is no dissipation (i.e. friction) and the flow is irrotational (up to quantum vortices). Theoretically it is described by a macroscopic wave-function with its absolute value related to the superfluid density and the gradient of its phase defines the superfluid velocity. In addition, in every superfluid there is also a physical Goldstone mode which costs nothing to excite.
Superconductor is like a superfluid, but the elementary particles are electrically charged (e.g. electrons). This leads to the Meissner effect, i.e. it costs energy for the magnetic field to penetrate into a superconductor. There is also no physical soft mode in the energy spectrum. Theoretically all this is explained by the Higgs mechanism.
In both cases the lowest energy quantum state is macroscopically populated which is know as Bose-Einstein condensation.