Close - the voltage at Q1 base will be 3.3V -Vbe . Vbe is typically 0.6V at room temperature, so the Q1 base voltage will be ~2.7V.
No - if the voltage at Q1 emitter is < Q1 base no current flows at all and the transistor is off. In fact Q1 E needs to be about 0.6V above Q1B (thats Vbe again) before any current flows.
No, that's the condition where Q1 turns on. A small current flows from Emitter to base, and a much larger current flows from Emitter to Collector. The ratio of emitter-base current to emitter-collector current is the gain of the transistor, also known as 'Hfe
This is quite true, except that the turn on voltage for Q1 would be 3.3V +Vbe, or about 3.9V. And it would vary roughly -2mV per degree centigrade.
I would recommend getting hold of a circuit simulator and trying it out. There are some good free ones, such as Simetrix (intro version), LTSpice, and TI-TINA. My preference is for Simetrix, which I've used extensively for many years - even did a small analog MOS chip design with it.