Ad 1) It refers to limit point in $X$. They do differ. Consider space
$$X=\left\{0\right\}\cup\left\{\frac{1}{n},\ \ n\in\mathbb{N}\right\}$$
and subset $Y=\left\{\frac{1}{n},\ \ n\in\mathbb{N}\right\}$. Obviously $Y$ has a limit point but it is not in $Y$. Furthere more every infinite subset $X$ has $0$ as its limit point. So the space is limit point compact in Munkres' sense but it is not limit point compact in your hypothetical sense.
Ad 2) Yes, at least under the Axiom of Choice. Let $E$ be an infinite subset of $X$. If it is infinite then you can pick sequence $(a_n)\subseteq E$ which is injective as a function (that actually follows from the Axiom of Choice).
If $X$ is sequentially compact then $(a_n)$ has a convergent subsequence $(a_{n_k})$. Pick one of the limits (note that the limit is unique if $X$ is $T_2$. We don't assume that here). Since it is injective then the limit must be different from almost all elements of $(a_{n_k})$. The limit might be one of the elements of $(a_{n_k})$ but not more due to injectivness. If you remove that problematic element from $(a_{n_k})$ you obtain a sequence with limit that is not a part of that sequence. By definition that limit is a limit point of $E$.
Ad 3) No. Consider the uncountable product of intervals $X=[0, 1]^{\mathcal{C}}$. It is well know that this space is compact but not sequentially compact. In particular there exists a sequence $(a_n)\subseteq X$ with no convergent subsequence. In particular $(a_n)$ treated as a subset of $X$ does not have a limit point in $X$. Otherwise we would be able to pick elements from $(a_n)$ that form a convergent sequence (again: under the Axiom of Choice).
It looks to me that this limit point compactness is really similar to sequential compactness (at least under the Axiom of Choice). Maybe be equivalent?
