Sort of and yes respectively.
When you're dealing with a resistive circuit, your first pass should try and condense the resistors down into an equivalent resistor. You can do this if the resistors are in simple parallel or series arrangements with each other but not with other circuit elements. We don't have that here.
Next, we start thinking about loops and junctions. If you trace your finger around a loop of wire in a circuit, say E2 -> R2 -> R3 -> E2, then we know the voltage changes over each of those elements has to add to zero -- after all, we need to end up at the same voltage we started at. If you identify a point where three sections of wire meet, such as point a, then if you add up the currents you have to get zero -- what goes in must go out. Collectively, these considerations are sometimes called Kirchoff's laws or Kirchoff's rules.
Together with Ohm's Law (V=IR, or voltage drop over a resistor is the product of current and resistance), this allows us to solve for currents and voltages everywhere in the circuit.
Along the top half of the circuit, we imagine the current going clockwise from E1 -> R1 -> R2 -> E2 -> E1. So:
E1 - I1R1 - I2R2 - E2 = 0
We write E2 as a negative rather than a positive because, as we go around the loop clockwise, we pass the "wrong" way over the second battery.
We can write a similar equation for the bottom half of the circuit. Since we already decided that the current in the middle wire is going to the left, we have to stick with that decision and imagine the current going counterclockwise around the bottom loop.
-E2 - I3R3 - I2R2 = 0
Also, we have all three strands of current meeting at a. We've imagine I1 going into a, I2 coming out and I3 going in. Currents going in are positive, coming out are negative.
I1 - I2 + I3 = 0
We now have three equations and three variables, so we can use whatever algebra we like to solve for it.
Note that it's ok to be wrong about the direction we guessed the current would go. We guessed that the current in the middle wire would go to the left, because we were going clockwise around the top half of the circuit and it would be convenient if the current were going to the left. If we are wrong about that, it means we'll get a negative number for I2. The magnitude of I2 will still be correct, we'll just have to note that it's actually going to the right and not the left
