How to write KCL and KVL in these circuits?
There's a voltage difference across the resistor R2. 

You can Omit \$V^{i}_{D2}\$because \$V^{i}_{D2}\$= 0.  

So, voltage across \$R_{2}\$ is =  0V - ( -10V) = 10V.

So, \$i^{i}_{D2} \$= 10V / \$R_{2}\$ = 10V / 5000 Ohms = 2 mA.

As you said: \$i\$ = 10V / \$R_{1}\$ = 1 mA.

You can substitute the values of   \$i^{i}_{D2} \$ and \$i\$ in the equation of KCL to get \$i^{i}_{D1} \$. That's because you don't have a resistor. So, I think this is the easiest way to calculate \$i^{i}_{D1} \$.

\$i^{i}_{D1} \$= \$i\$ - \$i^{i}_{D2} \$= -1 mA. (Negative means opposite direction).

You said \$i^{i}_{D2} \$= \$V^{i}_{D2}\$ / \$R_{2}\$ = 0 !! That's not true. Because \$V^{i}_{D2}\$ is not the voltage across R2. It is the voltage across the diode only (or the switch).

For the second question:

Edit: You should choose a smaller loop to include \$V^{i}_{D1}\$ such as:

\$-10V - V^{i}_{D1} + V^{i}_{D2} + i^{i}_{D2}  * R_{2} = 0 \$

Since there's no voltage across points A and B as you demonstrated in the picture, This means: \$  V^{i}_{D2} = 0 \$ and also \$  V^{i}_{D1} = 0 \$.

This makes the equation simpler: \$ -10V + i^{i}_{D2}\$ * \$ R_{2} = 0\$ 

or \$ i^{i}_{D2} = 10V / R_{2} = 0 \$

For the outer loop: Think of 10V and -10V as a battery of 20V and this battery feeds the two resistors so,

20V = \$i\$ * \$R_{1}\$+ \$i^{i}_{D1} \$ *\$R_{2}\$