It's digital logic, 3 being High and 1 being Low, anything in between won't have much effect.
The left side is an OR gate. So its kinda like High or low = 1 or 0 = High
High being 3 volts, and low being 2 or 1 volt relatively speaking.
In the latter, right side, is like an AND gate, So it's kinda like High and Low = 1 and 0 = 0
Again, High being 3 volts and Low being 1 volt.
I think the easiest method to solve such problems is to assume that the diodes are off (both, and then one of the two), compute the voltages across the diodes and see if there's a contradiction with your assumption. Let's call the top left diode \$D_1\$ and the diode in the middle \$D_2\$.
Case 1: \$D_1\$ off, \$D_2\$ off: Since \$D_1\$ is off there is no current through the top 5k resistor, and since \$D_2\$ is off, there is also no current through the bottom left 10k resistor. So \$V=0\$ and the voltage at the anode of \$D_1\$ is 15 Volts. Contradiction! (\$D_1\$ should be on).
Case 2: \$D_1\$ off, \$D_2\$ on: again no current through top 5k resistor. Voltage \$V\$ is
$$V=\frac{15V\cdot(5k||10k)}{10k+(5k||10k)}=3.75V$$
Contradiction! (Because the voltage across \$D_1\$ would be \$15V-3.75V=11.25V\$ and it should be on.)
Case 3: \$D_1\$ on, \$D_2\$ off: Voltage \$V\$ is
$$V=\frac{15V\cdot 10k}{5k+10k}=10V$$
The voltage at the anode of \$D_2\$ is \$15V\cdot 5k/15k=5V\$. This agrees with our assumption, because with these voltages \$D_2\$ must be off. So your solution is
$$I=0A,\quad V=10V$$
Best Answer
The method is just like any problem with an ideal diode.
Guess whether the diode is in forward or reverse bias.
Solve the circuit under that assumption.
Check whether the solution actually biases the diode the way you thought.
If it does not, assume the diode is biased the other way, solve again, and check your answer again.
For your circuits you'll have to essentially do this procedure for each switching event in the forcing function (for the \$t<0\$ initial state, for \$0<t<T\$, and for \$t>T\$). For the second circuit, you also have to consider the possibility that the diode changes state during the transient.
As you gain more experience you'll find it gets easier to guess correctly in the first step.