LM3915 to drive more LEDs / basic questions about PNP transistors in saturation
R1 is there to make sure T1 won't conduct if the 3915's output is off. When off there may still be a small leakage current. This won't do any harm if the output drives LEDs directly, but if you drive an external transistor with it the leakage current will be amplified, and the LEDs may light up very slightly, though maybe only visible in the dark. In any case, it's easy to prevent that. Imagine the 3915 slowly increasing its current. At for instance 10 &micro;A that current will go through R1, not T1's base. Why? 10 &micro;A through 1 k&Omega; is 10 mV, and that's far below the 0.7 V T1 needs to begin conducting. So even 100 &micro;A will only drop 100 mV, and still no base current. It's only when the current exceeds 700 &micro;A that T1 begins to conduct. Because the emitter-base voltage will be 0.7 V the current through R1 will be limited to that 700 &micro;A, what the 3915 will draw more comes from T1's base. In hindsight R1 may well be increased to 10 k&Omega;, then T1 will begin to conduct at 70 &micro;A, which will well be above the 3915's leakage current.

Since the 3915's outputs are current sinks R2 is not really necessary to limit the current. But without R2 almost the full power supply voltage (minus 0.7 V) will be across the output, and then the device's dissipation may become too high. Hence R2, to take some of that voltage, and the power which goes with it.  

You'll have to dimension R2 such that \$I_B \times h_{FE}\$ > required \$I_C\$, like you say. Note that for a transistor in saturation \$h_{FE}\$ is much lower than the value specified in the datasheet, so a base current of a couple tens of mA is no luxe. Check the transistor's datasheet for the maximum allowed base current.