Transistor operating point problem doesn't make sense?
This problem isn't a good candidate for hand solution.

Because the resistor divider formed by R2 and R3 is going to have about \$(10\ V)\frac{0.6}{9.3+0.6}\approx 0.6\ V\$ across R2, and thus there is about 0.6 V across the b-e junction of T1. 

That means that T1 is only very weakly turned on, and our basic rules for hand calculations don't handle this situation well. If we have a rule that the b-e junction must be -0.7 V (negative because it's a PNP) to turn the transistor on, then we think this transistor is in cut off mode. Or if we think the b-e junction needs -0.6 V to turn on then we see that the transistor is just barely turned on, but there's nothing to tell us how much current is flowing out of the base. 

Even if we use the more accurate 9.393 V you calculated for the Thevenin equivalent voltage of the 10 V source and resistor divider, there's nothing to say that the b-e voltage of the transistor will be exactly -0.6 or -0.7 under these bias conditions. It might be -0.65 V or -0.55 V, and given the small difference between the Thevenin source output and the base voltage in the circuit, this will cause substantial errors in the calculation result.

The best way to solve this circuit would be to plug it in to a simulator (with reasonably chosen transistor models). Or to adjust the resistor values to be sure the transistor is either fully in active mode or fully cut off.