Here goes...
Your signal has an output impedance of 1k. Therefore if you hang a resistor to ground on the end of this, you will get a voltage drop. Example: your output impedance is 1k. Put a 1k load on this and your output voltage will now be half what you expect it to be.
Your input impedance to your amp looks like a resistor down to ground. Your 1k source resistor will cause a voltage drop at the input pin if the input has any input impedance.
Now for your output resistance. This will look like a series resistor in series with your load. if you have a 1k output resistance: If you have an o/p voltage of 1v and no load, you will measure 1v output. You put a load on the output and your voltage will drop to 0.5V.
Now for your question... which I have forgotten. Let me scroll down..:
Ah yes..
Your source impedance is 1k. your input is Rin. Your voltage drop is:
Rin / (Rin + 1k) and this has to equal 0.5dB (which is 0.944). I got this from:
dB = 20 log (Vin/Vout). therefore inverselog(-0.5/20) = 0.944. We use -0.5 cos it is a loss.
Working backwards, we can find Rin = 16857ohms
Likewise with the output, your load is 5k:
5k/(5k + Routput) = 0.944
Routput = 296.6ohms
Hope this helps!