To solve these kinds of circuits you have to make an assumption about the state of each diode (whether it is on or off) and solve the circuit based on that assumption. If in solving the circuit you arrive at a contradiction (either the diode has a nonzero current through it but you assumed no voltage across it, or the diode has no current through it but you assumed 0.7V across it) then your assumption was wrong.
This circuit has only one diode so there are only two possible solutions: the diode is on, or it is off.
First assume that the diode is off (i.e. that the current \$I_3\$ through it is 0). By KCL that means \$I_1 = I_4\$ (you are correct that \$I_2 = 0\$ in steady state). Similarly, by KCL \$I_0 = I_1\$. \$I_0\$ is flowing through the two resistors in series so it equals
$$I_0 = \frac{U_0}{R_1 + R_4} = \frac{3.5}{280 + 350} = 5.5\text{ mA}$$
Since \$I_0 = I_4\$ the voltage across \$R_4\$ is \$U_4 = I_4 \times R_4 = 5.5\text{ mA} \times 350 = 1.94\text{ V}\$. However, \$U_4 = U_3 > 0.7\text{ V}\$ so the diode would be on. This is a contradiction so the diode must not be off as assumed.
Now assume the diode is on (the voltage \$U_3\$ across it is 0.7V). \$U_4 = U_3\$ so $$I_4 = U_4/R_4 = 0.7/350 = 2\text{ mA}$$ By KVL \$U_0 = U_1 + U_3\$, so rearranging we have $$U_1 = U_0 - U_3 = 3.5 - 0.7 = 2.8\text{ V}$$ That means $$I_1 = U_1/R_1 = 10\text{ mA}$$ By KCL \$I_1 = I_3 + I_4\$, and rearranging we have $$I_3 = I_1 - I_4 = 10\text{ mA} - 2\text{ mA} = 8\text{ mA}$$ We have a nonzero current through the diode so there is no contradiction -- the diode is on.
You should be able to figure out the other variables (like \$I_5\$) from here.
For a supermesh enclosing both current sources you need to come up with three equations. One equation will be a KVL equation around a loop that encloses \$I_A\$ and \$I_B\$. I see only one loop that satisfies that requirement and has no current sources on the loop itself.
The second equation states the relationship between \$I_A\$, \$I_1\$, and \$I_3\$. The third equation similarly states the relationship between \$I_B\$ and mesh currents. These last two equations you should be able to write by inspection.
Best Answer
Thanks for the addition. Just follow these steps:
simulate this circuit – Schematic created using CircuitLab
At this point, you've only one unknown node voltage, \$V_x\$, which is now easily computed using nodal analysis (or by simple inspection.) With \$V_x\$ in hand, you can work out all the currents in resistors \$R_1\$ through \$R_4\$.
(The current in \$R_5\$ is obviously already given. \$R_5\$ can be simply shorted out if you prefer, as its only impact is upon the voltage across the current source and no one cares what that is, anyway.)
With all the currents in hand, you can now use KCL on the top-right corner node to work out the current and its direction in \$V_2\$. And then use KCL on the node just above \$i\$ to work out the magnitude and direction of \$i\$.