The best way to explain this is to take a look at some circuit diagrams, and do some math.
We know that in a parallel circuit, the voltage across all loads is the same. So we know the voltage across each resistor is 120V, so we can easily calculate the amperage at each resistor.
I=E/R = 120V/1000 ohms = 0.12 amps
Now that we know the current at each resistor, we can find the total current using the following formula.
It = I1 + I2 = 0.12A + 0.12A = 0.24A
We find total resistance in a parallel circuit like this.
Rt = R1 x R2 / (R1 + R2) = 1000ohms x 1000ohms / (1000ohms + 1000ohms) = 1000000 / 2000 = 500 ohms
So Voltage = 120V, Current = 0.24A, and Resistance = 500 ohms on each leg A and B of the of the circuit.
Now if we break (open) the neutral, we have changed the circuit from two 120V parallel circuits to one 240V series circuit. Now we have to change our values, and do some new calculations.
We can use our total resistance calculations from before, to determine the resistance values for each parallel section of the circuit. We can now say we have a 240V series circuit with two 500ohm resistors, so we'll calculate total resistance by adding the resistances. Lets simplify our diagram.
Rt = R1 + R2 = 500ohms + 500ohms = 1000ohms
Next lets calculate the total current.
It = Et / Rt = 240V / 1000ohms = 0.24A
Now we can find the voltage at each combined load.
V1 = It x R1 = 0.24A x 500 ohms = 120V
We can see with a balanced load, we'll still see 120V like normal. But what happens when we have an open nuetral, and an unbalanced load?
Rt1 = 1 / (1/1000 + 1/1000 + 1/1000) = 1 / 0.003 = 333.33ohms
Rt2 = 1 / (1/1000 + 1/1000) = 500ohms
Rt = Rt1 + Rt2 = 333.33ohms + 500 ohms = 833.33ohms
It = 240V / 833.33 = 0.29A
VRt1 = 0.29A x 333.33ohms = 96.6657V
VRt2 = 0.29A X 500 ohms = 145V
As you can see, if you measure hot to neutral on leg A you'll read 96.6657V, while leg B will measure 145V.
What you're seeing is the different characteristics of parallel vs series circuits. In a parallel circuit voltage is constant and amperage varies, however, in a series circuit amperage is constant and voltage varies.
Best Answer
What you refer to as a power surge will most likely be a voltage spike, i.e. short but significant increase in the line voltage. This increased voltage will in the most general case lead to increased electrical currents in your devices, and this may kill them.
Simplest example: a light bulb that is specified for 230V will generate more light at higher voltages (because higher voltage -> higher current -> higher temperature of the filament -> more light), but due to the higher temperature, the filament will break earlier (as early as "instantly" if enough voltage is applied). Other simple examples might be motors which might run faster at higher voltages but will also suffer from the increased currents.
Any devices involving semiconductor circuits (so, almost all nowadays, I guess) are also affected in the same way. Too much current -> too much heat -> broken device. It is quite possible that only a part of a device is affected and the consequences might vary from fancy noises to complete failures.
So yes, it is possible
That totally depends on the nature of the voltage spike and the device.
There's no way of telling this for sure. It might be the case that the dying fridge was causing yet another voltage spike that further damaged the microwave oven, or the events might be completely unrelated.