Electrical – Has the emitter resistor any effect in calculating the base current in a common emitter amplifier

They apparently set the Rc/Re ratio to 3 because they wanted a voltage gain of about 3. Presumably that was desirable for the role of this circuit. Basically the Rc/Re ratio is the voltage gain assuming the transistor's gain is "large". Large means that you can approximate the collector and emitter currents as being equal, which also means the base current is 0.

You are right in that the Rc/Re ratio also sets the fraction of the supply voltage that the output can swing. As a simplification, consider that the transistor can vary from open to short C to E. When open, the collector (output) voltage is Vcc. When short, it is Vcc out of the voltage divider formed by Rc and Re. When Rc/Re is large, the output when the transistor is fully on approaches 0, and the output can swing the whole 0 to Vcc range. When Re is a significant fraction of Re+Rc, then it eats up some of the output voltage when the transistor is on, and the output can't swing as low to ground.

For good linear operation, you want to leave about a volt or so across C-E. The lowest output voltage swing is therefore Vcc-1V divided by the resistor divider, plus the 1 V across the transistor:

  Vomin = [(Vcc - 1V) Re / (Re + Rc)] + 1V

At high Rc/Re ratios, this approaches the 1 V you leave across the transistor. For lower ratios, the output voltage swing "lost" to Re is significant. All this is to say, yes, you're right, the output voltage swing depends on Rc/Re.

There is no rule of thumb for setting Rc/Re. This is basically the voltage gain of the amplifier. You set this to what it needs to be for other reasons.

However, you can't just make the gain infinite since then other factors that we can reasonably ignore at modest gains get in the way. These other factors are often hard to know. I'd say, try not to exceed a voltage gain of about 1/5 the transistor current gain. That's of course a tradeoff I picked out of the air, but with a reasonably good transistor gain, like 50 or more, a voltage gain of 10 is doable. Beyond that, the approximation that the transistor gain is "large" and you can mostly ignore the base current becomes less valid. So does the approximation that the B-E voltage is fixed.

As you can see at high gains the exact parameters of the transistor matter more and more. Since transistor gain varies widely, we generally want circuits to work with some minimum gain, but not rely on any maximum gain. Put another way, we want circuits to work with transistor gain from some minimum to infinity. The higher the gain, the more sensitive the circuit is to the transistor's gain and other parameters.

As a exercise, see what happens in your circuit when Rc/Re = 3 and the transistor gain is 50 (a quite reasonable minimum guaranteed value for a small signal transistor). Then analyze it again with infinite gain. You'll see only a rather small difference. Now do the same with a gain of 30, and you'll see much more sensitivity to the transistor gain.

It seems like you assume that you always need to know (and take into account) the Early voltage to calculate the output resistance.

OK, then I will tell you the Early voltage, let's say it is -100 V. Can you now do the calculations and determine Rout ?

If you do the calculation properly you will not get a significantly different answer ! Why is that ?

Look at the output and notice how Rc sits there. How does this Rc relate to the output impedance ? The output impedance can never be higher than the value of Rc.

In the question it is assumed that the output impedance of transistor (which is determined by the Early voltage) is significantly higher than the value of Rc. If the output impedance of the transistor was significant something would have to be specified, like the Early voltage.

Also, since the Early voltage is not even specified, how could you take it into account ? You cannot so you can safely assume that you can ignore the effect that the Early voltage has.