Re 1: yes, the exponential map can be defined for any Lie group as follows: we take an element $X$ of the algebra to the $g(1)$, where $t\mapsto g(t)$ is the 1-parametric subgroup such that $g'(0)=X$. This makes sense for any Lie group and depends only on the Lie group structure.

Re 3: any finite dimensional Lie algebra is the Lie algebra of a Lie group. This is a classical theorem on Lie groups, but unfortunately, the only proof I know of proceeds via the classification of semi-simple Lie groups and algebras. The exponential is a local isomorphism, so a sufficiently small neighborhood of the unit is in the image of the exponential map, and so every element of the group is a product of exponentials. Moreover, the Lie group can be assumed simply-connected (or else take the universal cover). Then a faithful representation of the algebra integrates to a locally faithful representation of the simply-connected group.

So the only thing that can prevent the answer to 4. from being "yes" (if I understand the question correctly) is that the image of a map of Lie groups may not be a Lie group. Off hand I'm not sure that a minimal-dimensional representation would do the trick but some representation would, as follows from Greg Kuperberg's answer here Is every finite-dimensional Lie algebra the Lie algebra of a closed linear Lie group?.

Re 1: yes, the exponential map can be defined for any Lie group as follows: we take an element $X$ of the algebra to the $g(1)$, where $t\mapsto g(t)$ is the 1-parametric subgroup such that $g'(0)=X$. This makes sense for any Lie group and depends only on the Lie group structure.

Re 3: any finite dimensional Lie algebra is the Lie algebra of a Lie group. This is a classical theorem on Lie groups, but unfortunately, the only proof I know of proceeds via the classification of semi-simple Lie groups and algebras. The exponential is a local isomorphism, so a sufficiently small neighborhood of the unit is in the image of the exponential map, and so every element of the group is a product of exponentials. Moreover, the Lie group can be assumed simply-connected (or else take the universal cover). Then a faithful representation of the algebra integrates to a locally faithful representation of the simply-connected group.

So the only thing that can prevent the answer to 4. from being "yes" (if I understand the question correctly) is that the image of a map of Lie groups may not be a Lie group. Off hand I'm not sure that a minimal-dimensional representation would do the trick but some representation would, as follows from Greg Kuperberg's answer here Is every finite-dimensional Lie algebra the Lie algebra of a closed linear Lie group?.