You should see immediately that the 7.09V can't be right. 7.09V on one end of the resistors and 12V on the other end can never give you 7V on the non-inverting input. Your equation for \$V_{REF}\$ seems to be wrong.
Here's how I do it. Since the current through the resistors is the same we have
\$ \begin{cases} \dfrac{V_{SAT+} - V_{UT}}{nR} = \dfrac{V_{UT} - V_{REF}}{R} \\ \\ \dfrac{V_{SAT-} - V_{LT}}{nR} = \dfrac{V_{LT} - V_{REF}}{R} \end{cases} \$
Filling in the parameters and eliminating R:
\$ \begin{cases} \dfrac{12V - 7V}{n} = 7V - V_{REF} \\ \\ \dfrac{0V - 6V}{n} = 6V - V_{REF} \end{cases} \$
From the second equation:
\$ V_{REF} = \dfrac{n + 1}{n} 6V \$
Replacing \$V_{REF}\$ in the other equation:
\$ \dfrac{12V - 7V}{n} = 7V - \dfrac{n + 1}{n} 6V \$
We find \$n =11\$, that's also the value you found. But my \$V_{REF}\$ is different:
\$ V_{REF} = \dfrac{n + 1}{n} 6V = 6.55V \$
This is OK for a theoretical calculation, but in practice you may have a problem: do you have a 6.55V source? The typical way to solve this is to get a reference voltage via a resistor divider from your 12V power supply, and then you get this circuit:
We still have 2 equations, but three unknowns, so we can choose 1. Let's pick a 30k\$\Omega\$ for R3. Then, applying KCL:
\$ \begin{cases} \dfrac{12V - V_{UT}}{R1} + \dfrac{V_{SAT+} - V_{UT}}{R3} = \dfrac{V_{UT}}{R2} \\ \\ \dfrac{12V - V_{LT}}{R1} + \dfrac{V_{SAT-} - V_{LT}}{R3} = \dfrac{V_{LT}}{R2} \end{cases} \$
Filling in our parameters:
\$ \begin{cases} \dfrac{12V - 7V}{R1} + \dfrac{12V - 7V}{30k\Omega} = \dfrac{7V}{R2} \\ \\ \dfrac{12V - 6V}{R1} + \dfrac{0V - 6V}{30k\Omega} = \dfrac{6V}{R2} \end{cases} \$
That's
\$ \begin{cases} \dfrac{5V}{R1} + \dfrac{5V}{30k\Omega} = \dfrac{7V}{R2} \\ \\ \dfrac{6V}{R1} - \dfrac{6V}{30k\Omega} = \dfrac{6V}{R2} \end{cases} \$
From which we find
\$ \begin{cases} R1 = 5k\Omega \\ R2 = 6k\Omega \end{cases} \$
The comparator output will change states based upon when the signal voltage at the + input goes above or below the reference voltage on the - input. Selecting the reference voltage has everything to do with the nature of the input signal levels. The sample circuit you show sets the reference at 2.5 volts. So you would expect the output to reflect a high whenever the input signal voltage is above 2.5V and low when the signal voltage is below 2.5V.
From this explanation you should be able to extrapolate the behavior of the comparator if you adjust the reference relative to a known input signal.
Do be aware that the comparator will have a valid operating voltage range for its inputs so do be aware of this as you go about changing the voltage on the reference and the input signal pin.
Best Answer
The emitter should be connected (via a resistor or not) to ground and not 5 volts: -
And, the collector normally has a pull-up resistor to 5 volts and not the complex arrangement you have shown (that doesn't match your other schematic).
So, forget about the emitter resistor and connect "E" to ground/0 volts. Use a simple 1 kΩ pull-up resistor from "C" to 5 volts and, feedback to "REF" with a high value resistor such as 100 kΩ or greater.
Here's an example from the data sheet that shows how E and C pins are used (my bits in red): -