Stable gain nyquist plot
There is a mistake in the last step. When you invert both sides of the inequality, the $&gt;$ inverts as well. So, for $K&gt;0$ (as you found that $K&gt;2$, so it must be positive):

$$\frac 1 K &gt; y \iff K&lt;\frac 1 y$$

If you do this when $y\to 0$, then you get the desired result.

Also, note that $\frac 1 K &gt;0$ would hold for any $K&gt;0$, as the only way to get a negative number from $1/K$ is letting $K$ be negative.

Regarding your other question, it would completely depend on the transfer function and its respective Nyquist. But don't worry, the procedure is exactly the same. For example, try doing the Nyquist plot for $-\frac{K}{1-s}$ and see what you get.