Given that \$\alpha\$ and \$\beta\$ are related by \$\alpha = \frac{\beta}{1+\beta}\$ as stated in the wiki article, obviously you can do your sums with either.
However, which is going to be easier to use? I personally always use \$\beta\$, regardless of the transistor configuration.
In common emitter \$I_c = \beta\times I_b\$, so I can say 'I need to control \$I_c\$ collector current, I need at least \$\frac{I_c}{\beta}\$ of base current to do it'.
But as \$\beta >> 1\$ (for most transistors), \$\alpha \approx 1\$, and \$I_c \approx I_e\$. You may object to the approximation, but given the way that \$\beta\$ varies with temperature, \$I_c\$, and between transistors of the same type, that is a far far better approximation than insisting that \$\beta\$ is constant. Any good transistor design will allow for operation with a range of \$\beta\$, at least \$2:1\$, preferably more.
Once you have made the approximation \$I_c \approx I_e\$, then common collector operation is given by 'I need to allow for a base current of \$\frac{I_c}{\beta}\$ to flow in the base circuit, without upsetting operation'.
With a common base stage, you say much the same thing, allowing an amount of base current, however you also say that the emitter to collector gain is slightly less than \$1\$, a fraction of \$\frac{1}{\beta}\$ less than one. The error of the gain from \$1\$ will usually be a smaller error than resistor tolerances and other sources of gain error.
Given that you can write an equation for \$\alpha\$, does that mean that you need to? For most practical engineering designs, the answer is no. If you are in college, and the tutor really likes to use \$\alpha\$, then the answer is yes.
You're correct, in this case $$I_B \neq 0$$
$$I_C = I_B +I_E$$
Instead, you're going to want to use a few DC relations for an assumed active BJT. I'm also assuming that you're given a value for Vt. You're going to want to make a KVL loop with IE and IB and Vt. It should look like something like
$$9V = I_E R_E + V_t + I_B R_B$$
with the added knowledge of
$$I_B = \frac{I_E}{\beta +1}$$
is enough to solve for IE. Use
$$I_C =\alpha I_E$$
to get the final current value. All thats left is to find the voltages using V=IR.
For the second part, of the question you'll have some algebra, but nothing overly complicated.
Best Answer
Yes you have assumed this, but this is still not a bad first stab analysis technique. For example, by assuming the BJT is in its active region of operation, you can do easy back of the envelope calculations to determine parameters like the voltage drop between collector and emitter to easily tell by contradiction if the transistor could not possibly be in its active region.
It is worth pointing out however that the inverse of this technique is not strictly true, of course. In practical engineering, the datasheet of a transistor will have guidelines on how to ensure reliable active operation of the transistor part over its rated environmental tolerances.