We know that standing waves are created when any wave traveling along the medium will reflect back when they reach the end. But in an open organ pipe, there is nothing to oppose the wave and reflect it back. So then how are standing waves created in such a case?
In standing waves where is the pressure greater, at nodes or anti-nodes?
Your first question is answered by this Open University article. To summarise, consider a low pressure region travelling along the tube towards the open end. The air outside is at atmospheric pressure, so when the low pressure region hits the end of the tube air from the atmosphere rushes in and creates a compression wave heading back down the tube. The opposite happens when a high pressure region hits the end of the tube.
The reflection of a sound wave at the end of a tube is an example of an impedance mismatch.
This also bears upon your second question. At an open end the wave inverts i.e. a reflected pressure peak becomes a trough, and a trough becomes a peak. This is in contrast to the closed end where a pressure peak reflects as a peak. This means the pressure changes are lowest at the open end and highest at the closed end.
The reflection at the open end is not 100%: some energy escapes. Actually this is obvious, because if no energy could escape an organ pipe wouldn't make any sound. The sound we hear is the energy that has escaped from the open end.
In organ pipes,vibrational antinodes will be present at any open end and vibrational nodes will be present at any closed end. When one plays a organ pipe, the air that is pushed through the pipe vibrates and standing waves are formed. If the both ends of the organ are open then the pattern for the fundamental frequency (the lowest frequency and longest wavelength pattern) will have antinodes at the two open ends and a single node in between.