Question on DC analysis of BJT – Common Base

What you see in the the textbook is perfectly correct. I think the confusion comes from the fact that the ratio \$i_b/i_{in}\$ is not spelled out in a very obvious form. \$i_b\$ is clearly in the numerator, but \$i_{in}\$ is in the denominator and must stay there. I made a kind of 'slow motion calculation' to show, how the ratios are rerranged:

\$\dfrac{\dfrac{\beta R_C}{R_C+R_L}i_b}{i_{in}}=\dfrac{\dfrac{\beta R_C}{R_C+R_L}}{i_{in}}i_b=\dfrac{\beta R_C}{R_C+R_L}i_b\dfrac{1}{i_{in}}=\dfrac{\beta R_C}{R_C+R_L}\dfrac{i_b}{i_{in}}\$

Afterwards, \$i_b/i_{in}\$ is simply available by applying the formula for the current divider:

\$ I_{BRANCH 1} = I_{TOTAL} \times \dfrac{Z_{BRANCH 2}}{Z_{BRANCH 1} + Z_{BRANCH 2}} \$

which clearly yields:

\$\dfrac{R_B}{R_B+r_{be}}\$

Does that answer your question?

In Nodal analysis (KCL),

Kirchhoff's Current Law implies that the sum of currents going in equal to the sum of currents going out. $$\Sigma_{k=1}^n {I}_{in} = I_{out}$$

A.) So you have two ways of going at it either

1. assume all currents are going in to a node. Then construct an equation in which that the the addition of all currents is equal to zero
2. intuitively decide in each branch if the current is going out, or going in. Then construct an equation where the left-hand side is the addition of currents going into the node, and right-hand side is the addition of the currents going out (or vice-versa)

B.) There exists a useful equation for each node in which more than two branches meet. So in this case there is 1 good equations. For each node i'd recommend assigning a variable (like Va for the one between all resistors)

We subtract the voltage at the tail of the direction of current to that of the head it would be as follows

(going into node) $$I_{1} = (30-Va)/8$$ (going out of node) $$I_{2} = Va/3$$ (going out of node) $$I_{3} = Va/6$$
note: current subscript correspond to resistor subscript $$I_{1} = I_{2} + I_{3}$$ $$(30-Va)/8 = Va/3 + Va/6$$

Simplify this and you notice Va = 6, thus the book answers are correct.