Stuck on designing a simple CS Amplifier
You want 5W across \$8\Omega\$ speaker. So we need \$I_L = \sqrt \frac{P}{R} = 0.790\$mA of RMS current or 1.2A peak current. This correlates with a voltage drop across the speaker 9V peak. So we need \$V_{DD}\$ larger then 18V but you choose 25V.

Next, we need to select the \$R_D\$ resistor value. 
For this type of a circuit, the maximum positive current at load is equal to: 

\$IL_{max} = I_{D} \times \frac{R_D}{R_D + R_L}\$ (1) 

And additional we want 

\$I_{D} = \frac{0.5 V_{DD}}{R_D} \$ (2)

And now we can solve for \$I_D\$ current (quiescent current) and \$R_D\$. 

$$I_D = \frac{V_{DD} \times IL_{max}}{V_{DD} - 2IL_{max}\times R_L} \approx 5.2\text{A}$$ 
$$R_D =  \frac{0.5 \left(V_{DD} - 2IL_{max} R_L\right)}{IL_{max}} - R_L \approx 2.2\Omega$$

But I do not like the results and decide to pick \$ R_D = 4.7\Omega\$ and \$I_D = 1.2\text{A}  \frac{8\Omega + 4.7\Omega}{4.7\Omega} = 3.25\$A

The power dissipation in \$R_D\$ resistor is \$50W\$ so we need a very big resistor.

And the power dissipation in MOSFET will also be equal to 33W. So you need a big heatsink as well. Because the efficiency of this amplifier topology is very low (6.25%).  

Of course, we need a capacitor in series with the speaker.

\$C_{OUT} \approx \frac{0.16}{F*R} \approx \frac{0.16}{20\text{Hz}*(8Ω + 4.7Ω)} \approx 680μ\text{F}\$ 

Now we need to choose a voltage divider resistors at the gate. 

This type of biasing network (without \$R_S\$ resistor) is very poor and should never be used in real life. Why? 

Because of the MOSFET \$V_{GS}\$ variation. 
Every transistor will have a slightly different \$V_{GS}\$ value at the same drain current. Also, we have a big power dissipation in a MOSFET hence the \$V_{GS}\$ will change with the temperature (Vgs will drop with the temperature and increases the drain current). So there's a chance of a thermal runaway.

As for your small-signal analyze is a way to complicated. 
The stage gain will be equal to 

\$A_V = gm \times R_D||R_L\$

The input impedance will be \$ R_{IN} = R_1||R_2\$ 

And the capacitors are just AC coupling capacitors (high pass). 
The value of a coupling capacitor is calculated with the lowest frequency you want to amplify (half-power gain -3dB). 
And for the input cap \$C = \frac{1}{2 \pi F * R} = \frac{1}{2 \pi 20\text{Hz} * R_{IN}}\$  

But ok, you should never build this amplifier in the real world. 

MOSFET Biasing Problems

The better topology for Class A amplifier is this one 







simulate this circuit &ndash; Schematic created using CircuitLab