Your mentor on this thread Audioguru seems to have taken a leave of absence for the time being. So let's see what I can do as a sub
Your main point of confusion seems to be in calculating the value of R1, specifically about the voltage across it. (You already know ohm's law). R1 and the 3rd R are in series and the same current flows from +9V to 0V. That is, the total voltage across the series combination of the two resistors is 9V.
Now, you've already chosen to have 2.62V across the 3rd R. That leaves (9 - 2.62)V = 6.38V (not 9V) across R1. Since the same 100uA flows through R1, you should now be able to calculate the value for R1.
If you arrive at the correct value for R1, that will place the base at 2.62V and the emitter at 2V. Since Re is 2k, that fixes Ie at 2V/2k = 1mA. That in turn fixes the collector current Ic at 1mA.
Ic flows through RL so that there's a voltage drop of 1mA*3.5k = 3.5V. Dropping 3.5V from 9V places the collector at 5.5V.
The collector-to-emitter voltage Vce = Vc - Ve. It's vital that Vce be at least a few volts so that the transistor can operate properly. I've left some of the numerical calculations to you as it will be good practice.
Note: The currents through R1 and the 3rd R are not exactly the same, and the same goes for Ie and Ic. The (small) difference in each case is the base current Ib. But taking them as the same is a close enough approximation for most applications like this. So let's not go into that for now. Maybe later, if necessary.