The input impedance is that seen by the circuit driving this circuit.
Figure 1. Input impedance is that seen between points 1 and 2.
In your case the driving circuit can only 'feel' the impedance between 1 and 2. As you correctly state the input impedance of the op-amp itself is extremely high in comparison with R1 and R2 and, as a result, doesn't load the input of the circuit to any significant extent. That leaves us with only R1 and R2 to worry about and they sum to 2k as is stated in the topic of study.
R4 presents itself as a load on the output of the op-amp (3) and doesn't affect the driving circuit.
Homework
Now, you should be able to work out the divider ratio for the input easily enough. If you want the gain of the whole circuit to be x 5 then what value would you assign to R4?
R4 should be 4k if I am not mistaken.
The op-amp with R3 and R4 form a non-inverting amplifier. The gain of these is given by \$ \frac {R_3 + R_4}{R3} = \frac {1k + 4k}{1k} = 5 \$ but you missed my gentle prompt that there is a 2:1 divider on the input so you need a gain of 10 on the amplifier. Oops!
What does R5 does for the circuit?
Nothing. It is, presumably, the load that the circuit is driving.
And if R2 was not connected, would the input impedance be infinite or 1K? Since there is no path to ground but through opamp itself?
Infinite, as you suspect.
