There are applications that are difficult without using a PNP. For example, if you want to do a high-side switch (load grounded) of a positive voltage, a PNP makes it easy. You just connect the emitter to the plus voltage and the load between the collector and ground. Of course the base must be driven between the positive voltage and ground to control the PNP and that is often performed by an NPN.
If you used an NPN for high-side switch you would have to connect the load to the emitter in an emitter follower configuration and the high voltage of an emitter follower can can no higher than about 0.7V below the positive voltage.
No, not really. But it's easier if you have a PNP within reach
[edit]
If you make a current sink with NPN and opamp, then the current is a little more tricky to calculate, because you must know the HFE to acurate calculate it.
Possible workarounds (NPN not included):
Use fet in adition to the NPN of just use a fet as replacement.
Use another resistor connected to ground for making a usable feedback.
If you make a current sink with NPN and opamp, then the current is a little more tricky to calculate, because you must know the HFE to acurate calculate it.
Not true. The negative feedback and high open loop gain of the op amp swamps the value of HFE. You can just calculate the current from the value of the resistor and Vin as shown on the diagram.
Just to clarify: In an NPN current sink, the sink current is indeed α*Ie, where α=β/(β+1). The same is true of a PNP current source.
As Carl pointed out, the base current is generally negligible (swamped, as he put it).