What happens in the second reaction solution will depend on the concentration of the reagents. When the hydrogen iodide concentration is high, it can form iodine. This is not a reaction of HI with an oxidant. What happens is
2HI ---> H2 + I2
If the HCl concentration in your experiment was very low in the second aqueous solution then I would not expect the reaction to occur. But if you look at an old bottle of concentrated HI, it will always be very dark due to the fact that it contains HI3. I know this as I have used HI solution to convert alcohols into alkyl iodides in lab.
Let us think for a moment about solvent extraction of metals from nitric acid, one of the problems is that the nitrate concentration is not equal to the anayltical nitric acid concentration (by titration) this is due to the fact that the Ka of nitric acid is not infinity. In the same way HI does not have an infinite Ka value.
As the anayltical concentration of HI in a solution increases, as the solvent properties will change and as Ka even in a very dilute solution is not infinity then the concentration of HI (intact HI) will increase. Also the concentration of the HI in the head space above the liquid will also increase.
The thermodynamics of formation of HI from I2 and H2 suggests that it should break down in to its elements. While Max Bondenstein did his work in the gas phase at a much higher temperture, it may be worth looking at.
Sulfuric acid will not form chlorine gas when it reacts with potassium chloride, it will form hydrogen chloride only. I know this as I have done this type of reaction myself.
When I was a teenaged boy I heated sodium chloride and sodium hydrogen sulfate in a test tube, it formed hydrogen chloride gas and not chlorine gas.