What is the inverse Laplace transform of squared denominator term?
This is quite straightforward to solve. Either just use this table where you can directly find the corresponding result, or "derive" it yourself with very basic knowledge of the Laplace transform.

You should know that

$$\mathcal{L}\{u(t)\}=\frac{1}{s}\tag{1}$$

Integration is equivalent to multiplication with $1/s$, so integrating $(1)$ gives

$$\mathcal{L}\{t\cdot u(t)\}=\frac{1}{s^2}\tag{2}$$

Replacing $s$ by $s+a$ is equivalent to multiplication with $e^{-at}$. Consequently, we have

$$\mathcal{L}^{-1}\left\{\frac{2}{(s+2)^2}\right\}=2\,t\,e^{-2t} u(t)\tag{3}$$