In general source degeneration resistor "adds" a negative feedback to the circuit (current-series feedback). In this case, we sample the output current (\$I_D\$) and return a proportional voltage in series with the input (\$V_{GS} = V_G - I_D*R_S\$).
This type of a feedback increases \$Rin\$ and \$Rout\$. But notice that the MOSFET itself has a very large \$Rin =\infty\$, therefore \$Rin = R1||R2\$ remains unchanged.
The voltage gain also drops to \$Av = -\frac{R_D}{R_S + 1/gm} = -\frac{R_D||R_L}{\frac{1}{gm} +R_S||R_3} \$
This also improves linearity, because without \$R_S\$ voltage gain is \$gm*R_D\$ and as you should know \$gm\$ varies with drain current.
Because \$gm\$ is a function of drain current (\$I_D\$), the voltage gain will vary with signal swing and the voltage gain also. But if we add external source resistance \$R_S\$ we notice that the \$R_S\$ does not change with the signal swing (\$I_D\$ swing)so, the overall voltage gain is stabilized and is more linear.
For \$R_S >> 1/gm\rightarrow A_V\approx \frac{R_D}{R_S}\$
Now let us look at \$rout\$. If we are looking from the load perspective we can see two paths for a AC current to flow:
First through \$R_D\$ resistor.
And the second one through MOSFET channel -->\$R_S\$ into GND.
As you can see now \$R_S\$ resistor is in series with the MOSFET channel.
So, to find resistance seen from the drain terminal into the MOSFET we need to use a small-signal-model.
\$r_x = \frac{V_X}{I_X}\$ and because \$V_G = 0V\$ we have:
$$V_{GS} = -I_X*R_S $$
And from KVL we have
$$V_X = I_{ro}*ro+I_X*R_S$$
$$I_{ro} = I_X - gm*V_{GS}$$
$$V_X=\left ( I_X - \left (gm\left ( -I_X \right )R_S \right ) \right )ro + I_XR_S $$
And solve for \$I_X\$
$$I_X = \frac{V_X}{ R_S + ro + gm*R_S*ro} $$
And finally we have
$$r_x = R_S + ro + gm*R_S*ro = ro(1+gmR_S+\frac{R_S}{ro}) $$
$$r_x = ro*(1+gmR_S)+R_S $$
As you can see adding \$R_S\$ resistor increase the MOSFET resistance.
The \$ro\$ is boosted by a factor of \$(1+gm R_S)\$
So, the overall \$r_{out}\$ is equal to:
$$r_{out} = R_D||r_x $$
and because \$R_D<<r_x\$ we have \$r_{out} \approx R_D\$
Best Answer
Current gain for MOSFETs is never referred to because it is meaningless. The drain current is not controlled by the gate current. (It isn't in BJTs either). When used in a circuit with other components the current gain can be relevant but it depends almost completely upon those other components.
With your example adding source degeneration (Rs) the voltage gain will decrease and so will the current gain because more current will be required to drive that voltage into R1 and R2.