Electrical circuit with capacitors and transistor
This is called an astable multivibrator. It's a beautifully simple circuit that is very challenging to explain! If you're new to electricity then don't get too discouraged - this is tricky for even experienced people to understand. There are many explanations on the web. This one is pretty good.

I highly recommend simulating the circuit, and attempting to answer all your questions by probing the circuit in simulation. If you don't already have a simulation tool, try CircuitLab (which happens to be the built-in circuit tool on this site) or LTSpice. They're both relatively easy to get started and will easily handle this sort of circuit.

Now to your questions:


Consider D1 on, D2 off. T1 is on so its collector is at ground allowing C1 to charge via R3. Once it gets to about 0.6V, T2 switches on clamping the voltage on that side of the capacitor. T1 switches off at the same time so now C1 is charging in the opposite direction via D1. The 9V source, minus the two diode drops in D1 and T2, means the voltage across C1 tends towards about 7V. Before it gets there the circuit switches again, giving you about 6.8V on the cap.
Through C1, as per answer above.
When T1 is off, there is a path from 9V through R2, C2 and T2, until C2 charges. There is very little drop across T2, so during that period V(R2) + V(C2) = 9V.
D2's turn off is caused by C1 - when C2 charges enough to turn on T1, C1 is suddenly pulled to ground which pulls the gate of T2 low, turning off D2.