Input resistance of a circuit containing op amps
I am trying to find the input resistance of the circuit itself, is it just simply the input voltage V1 minus the calculated voltage at the negative terminal of the upper left op-amp divided by 1k?


Check yourself with dimensional analysis. If you divide a voltage by a resistance, do you get a resistance, or something else?

The input resistance of a circuit is how much the voltage has to change to cause a change in input current:

$$R_{\mathrm{in}}=\frac{\mathrm{d}V_{\mathrm{in}}}{\mathrm{d}I_{\mathrm{in}}}$$

If you can't figure out this value by inspection, you can calculate the input current for two different input voltages and use the differences to find the derivative (since this is a linear circuit, it doesn't matter what two points you choose).

On your second question, 


  I am a bit stuck on how to force the input impedance seen by the two voltages to be 5k.


Think about what op-amp circuit gives a high input impedance and doesn't change the voltage signal. How could you modify it to have whatever input impedance you want?