R in and R out mosfet calrification
Let us first find the \$R_{IN}\$ for this circuit with the help of a T-model: 



As you can see \$R_{IN} = \frac{V_X}{I_X}\$ 

But we can not say that \$R_{IN}\$ is equal to \$ (R_g||1/g_m)+R_s\$ because this ignores the fact that \$g_m V_{gs}\$ flows into \$1/g_m\$ "resistance". 

This \$(R_g||1/g_m)+R_s\$ can only tell us the voltage across \$1/g_m\$ "resistor". 

But this will be easier to see in \$\pi\$-model 







simulate this circuit &ndash; Schematic created using CircuitLab

Now we see that all the \$I_X\$ current is flowing thought \$R_G\$ resistor. And \$I_S = I_X + I_D\$

So let us try to find \$R_{IN} = \frac{V_X}{I_X}\$  by applying the Kirchhoff's voltage law at the input loop. 

\$V_X = R_GI_X + (I_X + I_D)R_S = R_GI_X + (I_X + g_m V_{GS})R_S \$

What is \$V_{GS}\$ in this case? 

Well I hope that you see that \$V_{GS} = I_X R_G\$ therefore, 

\$V_X = R_GI_X + (I_X + g_m I_X R_G)R_S = I_X( R_G + R_S + g_m R_G R_S)\$   

And finally 

\$R_{IN} = \frac{V_X}{I_X} = R_G + R_S + g_m R_G R_S = R_G(1 +g_m R_S) + R_S\$ 

So for your circuit \$R_{IN} = 3 \cdot R_G + 2\textrm{k}\Omega\$

As a side note. 

Notice that this multiplication factor (\$3\$) is equal to: 

\$ \frac{1}{1 - A_V} \$

Where \$ A_V \$  is the voltage gain without \$R_G\$ resistor. 

\$A_V = \frac{R_S}{R_S + \frac{1}{g_m}} = \frac{2\textrm{k}\Omega}{2\textrm{k}\Omega + \frac{1}{1\textrm{mS}}} = 0.66(6) \$ 

\$ \frac{1}{1 - 0.66(6)} = 3 \$

Thus, form the input voltage source (\$V_X\$) point of view, the \$R_G\$ resistor looks bigger by a factor of \$ \frac{1}{1 - A_V} \$ And this is what we sometimes call a bootstrapping because we have a positive feedback here.

As for the output resistance \$R_{OUT}\$ you need to solve this circuit: 







simulate this circuit

\$I_X = \frac{V_X}{R_G} + \frac{V_X}{R_S} - g_m (0V - V_X) = \frac{V_X}{R_G} + \frac{V_X}{R_S} + g_m V_X = V_X(g_m + \frac{1}{R_G} +\frac{1}{R_S} )  \$ 

\$I_X = V_X(g_m + \frac{1}{R_G} +\frac{1}{R_S}) = V_X (\frac{g_m R_G R_S}{R_G R_S}+ \frac{R_S}{R_G R_S} + \frac{R_G}{R_G R_S}) =V_X \left(\frac{g_m R_G R_S + R_S + R_G}{R_G R_S}\right) \$

I hope that you see why \$R_4\$ do not appears in this equation. The situation will look completely different if you would add \$ ro = \frac {1}{\lambda I_D}\$ resistor (channel length modulation effect).

\$ V_X = \frac{I_X R_G R_S}{R_G + R_S + g_m R_G R_S}\$  

therefore

\$R_{OUT} = \Large \frac{R_G R_S}{R_G(1 +g_m R_S) + R_S}\$