MOSFET amplifier

We have to find out \$g_{m2}, i_d, v_{d1}\$ and \$R_D\$.

Since \$R_{i2} = 1/g_{m2} = 50\Omega,\$ \$g_{m2} = 0.02\mho\$.

Since both transistors are biased at the same point, \$g_{m1} = g_{m2}\$.
$$i_d = g_{m1}v_i = 0.1 mA$$ now, $$v_{d1} = i_d\times 1/g_{m1}=i_d\times 50\Omega$$

Now the equation in solution manual makes sense. I think you can do the rest your own.

This isn't a strange result. The ideal constant current source in the source circuit can have any voltage across and produce the specified current. In fact, V2 is essentially irrelevant.

It appears that you've specified that the current source produces a 0.5A constant current. Since the current source is ideal, the voltage across it will be whatever it needs to be such that there is 0.5A through the attached circuit. It 'doesn't care' what your supply voltages are - not one bit.

But this isn't remotely physical so, the first thing you should is get rid of the ideal current source. Also, if you want voltage gain, you shouldn't be using a common-drain configuration, as you have drawn, since the voltage gain is less than 1.

Honestly, I think it's likely that you're in over your head on this. For voltage gain, you probably want a common-source configuration and, for sure, forget that ideal current source.