Amplifying current using a buffer
The current amplification (Ai) is defined as the output current divided by the input current. So to calulate Ai you need to find expressions for ii and io, and divide.

First assume Cin and Cout are a short at the signal frequency. This is an approximation for AC signals; Now there are only resistors left. The suppply is a kind of capacitor too, so its also considered a short.

The input current is split into a current lost in the bias setup (Rb in this case), and the current into the base. The impedance of the base is given by hie. Knowing this you can calculate how much of the input current makes it into the base:

$$i_b = {{i_i R_b} \over {h_{ie}+R_b}}$$

Rewriting this gives the input current:

$$i_i = {{R_b i_b + h_{ie} ib} \over {R_b}}$$

The transistor amplifies the base current by a factor hfe to give the collector current; Together with the base current itself this gives the emitter current:
$$i_e = {(1+h_{fe}) ib} $$

The emitter current is split between Re and the output chain (R1+RL). So you calculate the current through the load:

$$i_o = {{R_e i_e} \over {R_1+R_L+R_E}}$$

Now, once you have expressions for both ii and io, you divide them, and see what's left. 

$$ A_i = {i_o \over i_i} = {{R_e {(1+h_{fe}) ib}} \over {R_1+R_L+R_E}} \cdot 
{{R_b} \over {R_b i_b + h_{ie} ib}}
 = {{(R_L+R_1)(h_{fe}+1)R_b} \over {(R_L+R_E+R_1)(h_{ie}+R_b)}}$$

At this point ib and ie disappear from the equation. All that's left are numbers to fill in.

For example take: hfe=100, hie=1000 ohm, Rb=100 kohm and Re=200 ohm. Using these values gives a current amplification of 8. 

So the circuit may be useful (provided the values chosen result in a valid DC operating point, and the supply can accomodate the voltage range implied).