Electronic – the use for non-inverting input resistance in a negative feedback opamp

The input resistance is the ratio between the change in the input voltage and the input current (or just between the input voltage and current in case of linear systems). In your case, assuming the input is on the negative side of the opamp, the input current is \$V_{in}/R_2\$ (because of the virtual ground on the negative terminal). So the input resistance would be \$R_2\$. If \$V_{in}\$ is applied to the positive side, the input resistance will be close to infinity, since there is no input current.
As for the output resistance, it can be obtained by connecting a known load \$R_L\$, measuring the voltage on it \$V_L\$, and then calculate the simple voltage divider problem: \$V_L=V_{out}R_L/(R_L+R_o)\$, where \$R_o\$ is the output resistance, and \$V_{out}\$ is calculated as for ideal opamp.

You can verify it using the common techniques for ideal opamps:
1) Having negative feedback makes the voltages on both terminals equal
2) There is no current flowing into or out of the terminals.

So, if \$I\$ is the current flowing through \$R_i\$ and \$R_f\$, $$V_n=V_p$$ $$I=(V_n-V_i)/R_i=(V_o-V_n)/R_f$$ Rearranging: $$V_o=\frac{R_f}{R_i}(V_n-V_i)+V_n=\frac{R_f}{R_i}(V_p-V_i)+V_p = -\frac{R_f}{R_i}V_i+\frac{R_f+R_i}{R_i}V_p$$ Which is exactly your formula.