Exercise 3.2 Biasing a circuit with two cascaded NPNs

a)

An answer:
Original situation: $$\begin{array}{rcll}& I_{C2}& =\mathit{\beta fe}{I}_{B2}& \\ & & I_{B2}=I_{E1}& \\ & & I_{E1}=(\mathit{\beta fe}+1)I_{B1}& \\ & & I_{B1}=\frac{V_{\mathit{CC}}-V_{B1}}{R_x}& \\ & & V_{B1}\approx 1.2V& \\ & I_{C2}& =\mathit{\beta fe}(\mathit{\beta fe}+1)\frac{V_{\mathit{CC}}-1.2V}{R_x}& \end{array}$$

Modified situation:

$$\begin{array}{rcll}& I_{C2}& =\mathit{\beta fe}{I}_{B2}& \\ & & I_{B2}=I_{E1}& \\ & & I_{E1}=\frac{V_{\mathit{CC}}-V_{B1}}{R_x}& \\ & & V_{B1}\approx 1.2V& \\ & I_{C2}& =\mathit{\beta fe}\frac{V_{\mathit{CC}}-1.2V}{R_x}& \end{array}$$

It follows from these relations that $I_{C2}$ is changed quite a lot when applying the described circuit modification.