Is the system $y(n)=x(n^2)$ time invariant?
The system is obviously time-varying. However you make the following usual mistake in the first case of showing the response to the delayed input $x_2[n] = x[n-k]$ as follows.

Writing the output $y[n]$ in the formal manner: $$ y[n] = \mathcal{T}\{x[n]\} = x[n^2] $$ and denoting the shifted input as $$x_2[n] = x[n-k]$$ you would have the corresponding output $y_k[n]$ as:

$$y_k[n] = \mathcal{T}\{ x_2[n]\} = x_2[n^2] = x[n^2 -k] $$ which is defintely not equal to $y[n-k] = x[(n-k)^2]$ hence the system is time-varying. 

You should have defined the output as $$ x_2[n] = x[n-k] \implies x_2[n^2] = x[n^2 -k] $$ but you didn't. See that the equality sign between $x[n]$ and $x_2[n]$ enables you to operate on $n$ only, not the whole argument of both sides.