As shown, it is a poor design, primarily due to wasting PIC POWER, and not turning on T2 hard enough.
To turn on T2 hard enough to prevent it from self-heating (i.e. to saturate it), its base current should 1/10th of its collector current. Since its base is at ~0.8V lower than 12V, and the collector of T1 is saturated (~0.1V) above ground, R3= (12-0.8-0.1)/0.01 = 11.1/0.01 = 1110Ω, say a std. value of 1K1.
Now, the collector current of T1 is ~10mA, so its base current when the port pin is high should be about 1mA (to saturate it). If the PIC Vdd is 5V, R1 = (5-0.7)/0.001 = 4K3.
That leaves R2: its function is to make sure that T2 is off when T1 is off. It will never have more than ~0.8V across it, and it shouldn't steal much base current from T2 when T1 is on, say <1% of 10mA, R2 = 0.8/0.1mA = 8k. Actually, the 4K7 that was there originally would work fine...
Stare at a sim of the corrected circuit: Note the base currents of each transistor...
Note that 1mA from the PIC becomes 10mA becomes 100mA at the load.