I don't specifically recall "almost compact" in the literature, but it's quite natural as it relates to "almost Menger" and "almost Lindelof".

In general, relative compactness is **not** equivalent to a set's closure being compact (sometimes called precompact). For example, in the particular point topology on a countably-infinite set, the singleton containing the particular point is relatively compact because it's compact. But its closure is the whole space, which is not compact. However, the claim does hold true for regular spaces (shown I believe first by Arhangelskii), so we probably should restrict our attention there.

Let $R$ be a relatively almost compact subset of a regular space $X$. We will show that $\overline{R}$ is compact (not just almost compact). Let $\mathcal U$ be a cover of $\overline{R}$. For each $x\in\overline{R}$ pick $x\in U_x\in\mathcal U$. Apply regularity to obtain $x\in V_x\subseteq \overline{V_x}\subseteq U_x$. Then we may cover $X$ using $\{X\setminus\overline{R}\}\cup\{V_x:x\in\overline{R}\}$. Apply relatively almost compact to obtain finite $F\subseteq \overline R$ with $R\subseteq\bigcup\{\overline{V_{x}}:x\in F\}$. Since $\bigcup\{\overline{V_{x}}:x\in F\}$ is closed, $\overline R\subseteq \bigcup\{\overline{V_{x}}:x\in F\}\subseteq\bigcup\{U_x:x\in F\}$. Since $\{U_x:x\in F\}$ is a finite subset of $\mathcal U$, we've shown $\overline R$ to be compact.

On the other hand, with or without regularity, $\overline R$ compact implies $R$ is relatively compact and thus relatively almost compact. So for regular spaces, relative almost compactness is equivalent to relative compactness and precompactness.

EDIT: One more result for regular spaces. Given almost compact $R$, cover $R$ with $\mathcal U$, where $x\in V_x\subseteq\overline{V_x}\subseteq U_x\in\mathcal U$ for each $x\in R$. Take $F\subseteq R$ with $\{\overline{V_x}:x\in F\}$ covering $R$, and therefore $\{U_x:x\in F\}$ covers $R$, showing $R$ is compact. Thus almost compactness is equivalent to compactness in regular spaces.

Rings of Continuous Functions(see 6J pg. 95). They define a Tychonoff space to bealmost compactif it has a unique compactification. If do not know if this usage originates from their book. $\endgroup$Products of Nearly Compact Spaces, where they are called $H(i)$-spaces (Stone had before labelled the property simply $(A)$ in hisHereditarily Compact Spaces). They have been studied as bothweakly compactandalmost compactspaces before. I think the references and discussion on pg. 182 of Mathur'sSome weaker forms of compactnesswill answer your first question. $\endgroup$1more comment