Here's the chemistry answer. Any contamination of a substance will expand the temperature range for which that mixture is solid. This is commonly used in labs to determine contamination of solids, a powder is slowly heated, and the temp at which it melts is recorded. That temp is compared to the melting point of the pure substance, and the difference shows how much contamination is present. Remember the scene from Blow when peewee is testing the coke's purity? Same thing is at work here, except contamination raises the boiling point. So, if water boils at 100C, after you contaminate it with sugar that BP will climb higher. So the wort should boil at a higher temp than pure water. Now, once it's boiling, that's it, it's boiling. There should be no difference in water temperature between a slow boil and a rapid boil, because the water is busy changing states and all the heat energy being added is going toward vaporizing the water, not heating it. The reason the boil might be slow at 75% burner and rapid at 100% burner is because you've increased the rate of heat input. Those bubbles are pockets of vaporized water nucleating on the surface of the pot. The faster you input heat energy, the faster water is vaporized and thus more bubbles form. The temperature of the water stays the same though. In fact, it's possible to supply heat so quickly that you would form a complete vapor barrier between the water and the pot, at which point you'd see the bottom of the pot begin to overheat and glow red. This is a one of the things nuclear boiler operators concern themselves about. Anyway, point is, the temp of the water is the same no matter what the boil rate is, and that temp is higher than 100C if there's sugar dissolved. Chemical reactions in the wort shouldn't be affected by the speed of the boil, theoretically, but the mixing action of a fast boil may contribute something to it, preventing a gradient from forming or some jazz like that.
As far as why an electric stove has trouble getting those last few degrees of temp, it's just a matter of energy in - energy out = mass*Cp*change in temp. The energy in coming from the stove is maxed out, but the energy out increases as the temperature increases (the temperature gradient between the hot water and the surrounding air is higher, thus there is a larger driving force for the energy to go to the air). At a certain temperature the energy in - energy out will equal 0, and the temp will stop climbing, whether it's boiling or not. Ways to get the temperature to go above this bottleneck are using a lid (increases the pressure and reduces convective heat loss) or swapping from a metal to a ceramic pot (reduces the conductive heat loss). Or you could just do your boil in a room with an ambient temperature of 80C
