The offset with the zener is of little use. The best zeners have a 1 % tolerance, which is 100 mV for a 10 V zener. On a 10-bit ADC this is a 20 count error, on a 12-bit ADC an 82 count error. You could trim the error away if you can measure the voltage accurately enough, but there are other factors. The BZX84-A10 has an 8 mV/°C temperature coefficient, giving a 2 count error per °C change in temperature for the 10-bit ADC, and 7 counts for the 12-bit. Looks like it's better suited as thermometer than as voltage meter. When you use a 10 V zener you'll also need a higher voltage power supply.
The resistance divider will do much better. Resistors also have a tolerance, but at 25 ¢ a 0.1 % resistor is still affordable. (Better that 0.1 % becomes expensive quickly: a 0.05 % costs almost 1 dollar.) At 10-bit resolution that will give you a 1 count error, 4 counts at 12-bit. Temperature coefficient will be less of a problem if both resistors are from the same series and placed close to each other: since the divider is ratiometric resistance changes will cancel each other out.
The numbers indicate that a higher than 10-bit resolution is of little use; component tolerances and variations will cause extra bit to be unreliable. A few extra bits may help to increase noise immunity, though, by averaging a series of measurements, or using a sigma-delta ADC, which averages the input signal anyway.
There's also something more philosophical: we always want better, but why on earth would you want to know a 12 V battery's voltage to a precision of better than 10 mV. You'll have a hard time getting the required resolution, and you'll always be uncertain about that last digit.
The ADS1000 is a low cost 12-bit ADC which will operate from a single 5 V supply.
Differential voltage means exactly that - the difference of the two input lines (Vin+/Vin- and Iin+/Iin-). The CS5463 can measure up to 250 difference between them (or 50mV with a higher gain). Both signals are referenced to AGND, and can range from 0.25V below AGND up to Vref (max. 2.5V). This is stated on page 7 in the data sheet (the maximum rating are a little bit higher, see page 13).
Your first question is answered partly in the explanation of your example circuit (Figure 18 on page 41). The data sheet says In this type of shunt-resistor configuration, the common-mode level of the CS5463 must be referenced to the line side of the power line. (Though it doesn't state why). One reason might be that you then don't need to break the neutral wire (which should not be done, neutral should be as close to earth as possible).
If you revert the plug, nothing bad will happen to the circuit - as long as you isolate any outputs. When you interface with other circuits which might be ground-referenced, you can create short circuits. Depending on the cabling of you house and the appliance you want to measure, you might trigger a RCD. (I'm not an expert here, hopefully someone more knowledgeable can explain this better)
Your resistor calculation seems OK. The input voltages can actually be higher that 250mV, it is just that this is the maximum voltage you can measure (anything higher will read as 250mV). So if you want to be able to read higher voltages (where I live we have 230Vrms) or detect over-load conditions, you should leave a safety margin (50mV seems fine).
The filter are connected to ground because you have different kinds of noise. You might have noise only on one line, or on both lines. For the voltage input, I would size Rv- like R1, and for the current input significantly larger than Rshunt. Then the capacitors can be sized to have a cut-off frequency of several kilo-hertz, so you won't influence the measured signal. You might also want to look at the evaluation board schematics. There are examples for the values (though it's schematic looks a little bit different).
Best Answer
I)
It doesn't. Whether it's low or high, the resistors act as a pull-up while also providing the needed signal. The current, for a supposed logic low would be \$115\sqrt{2}/940k\Omega = 173 \mu A\$
As Some Hardware Guy said, it's floating so the reference is 0V because its supply is mono-alternance and done through a capacitive impedance (reactance) of \$1\mu F\ = 2.65k\Omega \$ (60Hz) or\$ 3.18k\Omega\$ (50Hz).
They're there for not actually putting the pin to 160V, your fear at #1.
II)
The rest:
The LED will still have ~2V foreard voltage drop and the maximum current will be: \$ I_{max}=\frac{115\sqrt{2}-2}{100k\Omega} = 1.62mA\$
There won't be -120V since its reference will be the IC's supply, therefore the bridge will clamp the voltage.
You're welcome, but you should really try those simple circuits that have proven to be good enough, instead of choosing this.