Emitter-Follower Class A amplifier
Simply, they use a Large-Signal analysis. 

The small-signal analysis should only be used to find voltage gain only. 

As for the \$V_{BE}\$ equation. It's strange that you never see this equation before in the book? Especially that is it in chapter 12. But ok.   

First, think you need to do is to rearrange the Shockley equation and solve it for \$V_{BE}\$. 

$$I_C = I_S \times e^{\frac{V_{BE}}{V_T}}$$ 

$$\frac{I_C}{I_S} = e^{\frac{V_{BE}}{V_T}}$$ 
$$\ln\left(\frac{I_C}{I_S}\right) =\frac{V_{BE}}{V_T}$$ 
$$V_T\ln\left(\frac{I_C}{I_S}\right) =V_{BE}$$ 
$$V_{BE} = V_T\ln\left(\frac{I_C}{I_S}\right)$$ 

Now let's define \$ΔV_{BE}\$ as the difference between the actual \$V_{BE1}\$ voltage and some reference \$V_{BE2}\$ voltage for a different collector current.

$$ΔV_{BE} = V_{BE1} - V_{BE2} = V_T\ln\left(\frac{I_{C1}}{I_S}\right) - V_T\ln\left(\frac{I_{C2}}{I_S}\right)=V_T \left(\ln\frac{I_{C1}}{I_S}- \ln\frac{I_{C2}}{I_S} \right) $$
$$ΔV_{BE} = V_T \ln\left(\frac{I_{C1}}{I_S} \times \frac{I_S}{I_{C2}}\right)=V_T \ln\left(\frac{I_{C1}}{I_{C2}}\right)$$
$$ΔV_{BE} = V_T \ln\left(\frac{I_{C1}}{I_{C2}}\right)$$

Therefore, if we know the \$V_{BE1}\$ at \$I_{C1} = 1mA\$ we can calculate the \$V_{BE2}\$ value at different current \$I_{C2}=4.8mA\$. 

\$V_{BE2} = V_{BE1} + ΔV_{BE} \$ 

The end of the story.