Electronic – How current can flow through open circuit output and 9v across diode

My, that's a lot of questions.

Presumably the diode model you instructor wants you to use is an ideal diode with 0.7V drop in series with a 4 ohm resistor. So, when the diode is conducting, it behaves like a voltage source with a resistor in series. When the forward bias is less than 0.7V it does not conduct.

Is the 4 ohms per diode negligible in comparison to 1.5K? Well, it's more than 0.5% for two diodes so it will drop tens of mV. That might be negligible or it might not be, depending on the application. Since the instructor gives you the value, I suggest it might not be negligible in terms of getting the correct answer.

Two such diodes in series behave like one 1.4V diode with 8 ohms in series.

Here's what the 1N4148 signal diode characteristic looks like in the foothills: -

As you can see, below ~0.8 volts it's a pretty regular shaped characteristic. The current is approximately falling by a factor of ten for each 100 mV reduction in base-emitter voltage.

On a linear graph it looks like there is a real definable knee voltage: -

But, that linear graph hides the reality/subtlety of things but who cares - most diode circuits are nicely approximated by saying current is zero below 0.6V (or 0.7V) and the diode behaves like a short at this voltage.

There is an equation called the Shockley diode equation that adequately describes current versus voltage based on temperature and reverse leakage - often a sim will use this type of equation but rough and ready calculations on a scrap of paper don't need to.