Currents will always flow towards the lowest potential (i.e. in the context of VCE physics, the lowest 'voltage'), which in most cases is the ground, since it's 0 V. Here though we have a voltage that is lower than the ground, that is -3 V, as in this case our ground is just a reference. For a, there is 5 V dropped across the first resistor, so I=V/R=5/5K=1 mA. For the second resistor there is 3 V dropped over the 2K resistor, so I=3/2k=1.5 mA.

As to the two different currents part, the second current is greater than the first current, that is because there will be some current leaking from the ground, as again, the current will flow towards the lower voltage. So from 0 V to -3 V, i.e. that is where your extra current is coming from.

For b, if we wanted to find the voltage dropped over each resistor, we would use a voltage divider, but we want the current through each, which is the same through each. So 8 V is dropped over 7 K worth of resistance, that is I=8/7K =1.14 mA.

As for the direction, the will all be flowing in the direction P, from higher potential to lower potential.

Hope that helps.