As cjl mentioned, this is simply not true. It is sometimes true, but not universally. For example, the F-104 Starfighter had a supersonic wing. Shoot, it even had a supersonic fuselage, thus the need for the inlet cones on its engines. In general, the flow behind an oblique/conical shock is still supersonic.
Even on a subsonic airplane you can get supersonic flow locally in the transonic regime. I should point out that this does not
necessarily lead to separation (though it certainly can), does not
necessarily lead to transition to turbulence and certainly does give rise to wave drag. Of course, on a supersonic aircraft, even if the wing was subsonic, there would still be wave drag because the nose is still creating a shock.
That isn't true. It depends on a number of things. The oblique shock arises from the fact that the flow has to turn, and in general, there are two β solutions (shock angles) to the so-called θ-β-M equations for a given incoming Mach number. The smaller solution is called the ordinary or weak solution, while the larger one is called the extraordinary or strong one. In almost every situation, it is the ordinary solution that forms, and for every
ordinary solution, the downstream Mach number is supersonic. You can have an oblique shock that is supersonic if your downstream pressure is high enough, leading to the formation of the shock governed by the extraordinary solution.
If you get slow enough or the deflection angle gets large enough, the two solutions approach one another and eventually no solution exists, signifying that the shock detaches, becoming a bow shock. In that case, you have a much more complicated situation and the flow immediately behind the normal region is subsonic but quickly accelerates back to supersonic speeds most of the time as it curves around the geometry.
Assuming the flow is just above Mach 1 so that a shock actually does form, the angle will not be 45° in general unless the turning angle of the flow is also infinitesimal. Otherwise it is a Mach wave. If we do assume that the turning angle is such that it generates a shock, the angle will not necessarily be 45° and the downstream Mach number will either be supersonic and the shock will be oblique, or else the shock will detach and be locally supersonic before accelerating again. For a small Mach number like M1
=1.1, you would need a turning angle on the order of 1° in order to have an attached shock, and then you will have β≈19° and M2
≈1.03, which is still supersonic.
Of course, as that air downstream of the shock is then expanded around the curved nose cone, it will undergo an isentropic expansion (ignoring the boundary layer of course) and you will see it speed up rather than slow down.
Again, that depends entirely on a number of parameters, including the nose LE radius of the wing, the Mach number approaching the wing and the geometry of the wing behind it. There is a reasonably good chance that in this case, the flow over the wing will still be supersonic. It may not be quite as supersonic as the free stream, but you cannot generally state it is subsonic. Generally speaking, modern supersonic aircraft are designed to avoid this. You can even look at the wing profiles and tell it is supersonic because they are generally long, thin and have very, very little camber since in a supersonic flow, thickness and camber contribute exactly zero to the lift, while having large drag implications.
Hypersonic airfoils follow the same general guidelines as supersonic airfoils. They simply have additional considerations when dealing with thermal management and drag, among other things.
It has nothign to do with "punching" through the sound barrier, which doesn't exist. Transonic airfoils designed to operate in that range are not even generally very thin. Look up supercritical airfoils. They are an entirely different concept. Supersonic airfoils are thin and uncambered because neither thickness nor camber contribute to supersonic lift, but they both contribute to supersonic drag.
EDIT: Just realized that I repeated a lot of what cjl said. Haha.