It does sound like you're over-thinking things a little, but that's probably because you're struggling to reconcile several seemingly contradictory principles in your head. Don't feel bad. This stuff is hard.
One thing I think you're getting caught up in is the difference between "applying a voltage across a diode" vs implicitly calculating the voltage across a diode. If all you had was a 3V power supply and a diode, then you'd be explicitly applying 3V across that diode:
In this circuit, you can't apply a voltage across the diode as you did above. The voltage at NODE1 is a function of the diode and the resistor. Like a balanced ecosystem, everything in a circuit can affect everything else.
So will the diode conduct? Assuming an ideal diode, you can answer that purely on observation without doing any math. The anode of the diode is directly connected to the highest potential point in the circuit: 3V. The cathode (NODE1) is connected to a resistor which is connected to ground. The diode is clearly positively biased and current will flow through it. Since this is an ideal diode, there is no voltage drop across the diode, so NODE1 will also be 3V. Now that we see 3V at NODE1, we can forget about the diode and just use Ohm's Law on the resistor. A 3V drop across a 1k\$\Omega\$ resistor will result in 3mA through the resistor.
Now let's introduce a circuit with two diodes and one resistor.
I like to approach circuits like this by imaging time slowing way down and watching what happens in super slow motion. Before the voltages were applied, NODE1 was resting at 0V, so we'll start there. The 2V and 3V supplies are attached to the inputs and current starts flowing. At that first instant, both diodes are positively biased: there's a positive voltage on their anodes (3V and 2V, respectively) and their cathodes are both at 0V. You can imagine those first couple of electrons starting to flow through both diodes. But that will cause the voltage at NODE1 to begin to rise.
Lets say a tiny fraction of a second later the voltage at NODE1 has raised to 1V. The anode of D1 is 3V and the cathode is 1V. Positively biased. The anode of D2 is 2V and the cathode is 1V. Also positively biased. Current still flowing through both diodes. Another tiny fraction of a second later, the voltage at NODE1 raises just above 2V. For D1, this is no problem, it's still positively biased. But now D2 has a slight higher voltage on its cathode than its anode. D2 suddenly becomes reversed biased and stops conducting. Now D2 can be ignored from the circuit as D1 continues to conduct. Eventually, the voltage of NODE1 will become 3V since D1 is the only diode that is still conducting.
All that happens in a timespan measured in millionths or billionths of a second. But it shows that both diodes do conduct at first. But eventually, the diode with the larger voltage on its anode dominates the diode with the lower voltage. You should be able to easily apply that same reasoning to a circuit with 3 diodes and a resistor.
You can use the same slow motion technique on problem E4.4(f) in your question. In that case, assume the voltage at the top of the resistor was 0V and then suddenly switched on to 5V. Initially, the anodes of all 3 diodes will be 0V and then begin to rise. Which diode will start to conduct first? Once that diode is conducting, will the voltage at the anodes continue to rise or will it stop there? If it continues to rise, what happens to the amount of current going through the resistor (use Ohm's Law)? How does that resistor affect whether the voltage continues to rise or stops at the first diode?
Use this schematic to test the diodes. You can easily distinguish Silicon and Germanium Diodes. Silicon diodes should read approx 0.7V and Germanium diodes should read 0.3V. A little difficult to distinguish Schottky diodes though. They should show approx 0.2V which is close to 0.3V. If you have a very stable power supply and a good meter you can distinguish this as well!