Question 1
R5, R6 and R7 combine to form a total resistance of about 1MΩ, a value needed for the transfer function which I describe below. However, a single resistor would probably not be able to withstand the hundreds of volts that the inputs are subjected to (by source V2). This series arrangement shares the total voltage equally across three indiviual resistors, to bring the voltage each resistor is exposed to down to acceptable levels.
U2 and its peripheral resistors form a classic differential amplifier, but with a DC voltage offset applied to COM1. Simplified a little it looks like this:
simulate this circuit – Schematic created using CircuitLab
The output of this circuit is:
$$ V_{OUT} = V_{COM1} \cdot (1+k_1) \cdot (1-k_2) + V_R \cdot k_2 \cdot (1+k_1) - V_N \cdot k_1 $$
where
$$ k_1 = \frac{R_2}{R_1},\ \ k_2 = \frac{R_4}{R_3+R_4}$$
Plugging in the resistances, we get approximately:
$$ V_{OUT} = V_{COM1} + 2.22 \times 10^{-3} \cdot (V_R - V_N) $$
Question 2
I see two possibilities for the purpose of R4. The first is to tie the differential inputs together in the absence of any source being measured, but this seems very unlikely because:
It's a single resistor subject to the same conditions that R5, R6
and R7, and would likely need to be split into 3 separate series
resistors for the same reason.
Even if the inputs were unconnected, the nature of this circuit is
such that the output would settle at its "zero input difference"
quiescent level. That would be equal to \$V_{COM1}\$.
That leaves the only other reason I can think of, which is to help the simulator work. Sometimes disconnected voltage sources in a simulation can throw errors.
Question 3
Given that the ADC expects an input between 0V and 3.3V input, as you say, and also that the opamp output cannot swing all the way to the positive 5V rail, I believe you are correct.
This opinion is supported by the transfer function of the differential amplifier, which clearly adds this 2.5V (\$V_{COM1}\$) offset to the output. Unless there's something we don't know, setting \$V_{COM1} = \frac{3.3V}{2}\$ makes more sense.
