Why do the tangent holes (like in pitot-static tube) feel the static pressure? Why do the tangent holes (like in pitot-static tube) feel the static pressure although the static pressure is in the direction of the velocity of the streamlines not normal to the surface of the hole
 A: There seems to be some confusion:
Stagnation Pressure is likely what you are thinking of - it is in the direction of fluid velocity, and includes the dynamic pressure + the static pressure.  The distinction is important.  
Stagnation pressure is a measure of the energy in the system.  You can't measure stagnation pressure with a tangent hole.  You use a pitot tube to measure that.
Static pressure is like water pressure from your garden hose, when you plug up the end with your thumb.  No water is flowing, but it pushes back with a lot of pressure.  The tangential hole, especially when the diameter is much smaller than the ducting you are measuring, is usually behind a boundary layer.  It can't measure any dynamic pressure, so it only measures the static pressure.
Dynamic pressure is energy due to the air moving.  It is measured by taking the difference between the stagnation pressure and the static pressure.  It uses a device exactly like the one you depict here, measuring the difference between the two pressures with a tangential hole and a pitot tube.
If you want more information, or just want help with design or principles of any of these kind of measurement devices, SE.Engineering can also help.
A: The answer, according to John Denker at av8n.com is that the decrease in static pressure due to velocity is, to a first approximation, $1/2\rho v^2$ and $v$ is small compared to the speed of sound.
It's small enough to neglect.
As he says:

As you increase your airspeed, the stagnation pressure goes up, but
  the static pressure does not.

A: Pressure is isotopic (it acts equally in all directions) so it doesn't matter how the static pressure hole is oriented (it just happens that a hole in the tube sidewall is convenient and hence is tangential to the flow). Nevertheless the pitot tube must face into the flow in order to bring it to rest. 
A: It's because the Prandtl condition is valid. The pressure on the streamline is the same as the pressure at the hole,
$$\frac{\mathrm dp}{\mathrm dy}=0$$
