OK, let's see what we can do about that last question.
For the moment, let's continue with our analogy of a mechanical switch and a transistor used as a switch. Imagine a switch which does not have a snap action, but instead has considerable resistance if you push it lightly and make better contact (lower resistance) when you push harder. At some point, the resistance will reach its minimum and no longer decrease even if you keep pushing harder. That minimum resistance condition is analogous to transistor saturation and the push force is the base current.
To simplify things for the next part of the explanation, omit the LED and have only a resistor (RL) as the collector load. If we start with a very small base current, it will cause a certain amount of collector current to flow (Ic = Ib*beta). That Ic will cause a voltage drop in RL. The voltage drop is Ic*RL and the remaining voltage is across the transistor's collector-emitter terminals.
For example, if we start with Ib of 20uA and beta is 100, Ic will be 100 times 20uA which is 2mA. If RL = 100 ohms, the voltage drop across it = 2mA*100 ohms = 2V. If Vcc =5V, after dropping 2V across RL, we'll have 3V remaining at the transistor collector. Since the emitter is at ground or 0V, Vce is 3V under that condition.
As we increase the base current, Ic will also increase until almost all the available voltage is dropped across RL. The only voltage remaining across the transistor will be the saturation voltage Vce(sat) which is beyond our control.
For the moment, let's go back to our hypothetical non-snap mechanical switch above. It will be very difficult to apply just enough pressure on the switch to reach minimum contact resistance. So what we'd normally do is to apply enough extra pressure to make sure that the contact is in minimum resistance condition.
In the same way, it's almost impossible to calculate what would be just enough base current to cause the whole supply voltage to drop across RL. So we apply enough extra base current to ensure that practically all the supply voltage is dropped by the load resistor RL and the only voltage remaining across the transistor c-e is Vce(sat).
So how do we know how much base current to apply? A BJT amplifies the base current by a certain factor. But that current amplification factor (beta) varies from unit to unit even for transistors of the same type number. It also changes with temperature, with Vce and with current levels. A particular transistor may have beta of 100 at Vce of 5V and Ic of 10mA. But beta for the same transistor will change at different voltage and current levels. It drops quite a lot as we lower Vce down to saturation levels.
The usual practice is to use Ic/10 as Ib for medium beta transistors like the 2N2222, and Ic/20 or Ic/30 for high-beta transistors like a BC547B. These ratios are rule-of-thumb figures, not precisely calculated ones, just as we would apply an estimated amount of pressure on the non-snap switch.
Now let's start calculating practical values for your circuit:
We've already estimated that, at Ic = 20mA, Vce(sat) for a 2N2222 is about 50mV or 0.05V. VLED = 3.5V, making a sub-total of 3.55V. That leaves (5 - 3.55)V or 1.45V to be dropped by the resistor.
R = V/I = 1.45V/20mA = 72.5 ohms. The nearest standard value is 68 ohms so that the actual current will be 1.45/68 = 0.021A = 21mA.
Using our rule-of-thumb formula, Ib should be about 21mA/10 or 2.1mA.
Again estimating Vbe as about 0.7V, the base resistor has to drop 4.3V at a current of 2.1mA.
4.3V/2.1mA = 2.05k. The nearest common value is 1.8k or 2.2k.
Whew! I wanted to make the reasoning behind the calculations understandable and knew that it would take a fairly long post. That's why I didn't tackle it at that late hour last night. If something's still not clear enough, just ask.