Let us consider for simplicity the complex semi-simple case. As you mention, every irreducible representation $\rho$ of a semi-simple finite dimensional Lie algebra $g$ integrates to a representation of the corresponding simply connected Lie group $G$. But even if we suppose $\ker\rho=0$, the representation of $G$ may well have a kernel, a finite central subgroup $H$ of $G$. So we obtain representations of $G/H'$ where $H'$ is a finite central subgroup contained in $H$, but not of other groups with Lie algebra $g$.
So one could ask: given a Lie group $G'$ with Lie algebra $g$, when does a representation $\rho$ of $g$ with highest weight $\Lambda$ integrates to a representation of $G'$? The answer: if and only if $\Lambda$ belongs to the character lattice of a maximal torus $T$ of $G$. (Recall that the character lattice of $T$ is the discrete additive subgroup of the dual of the Lie algebra of $T$ spanned by the differentials of the homomorphisms $T\to\mathbf{C}^{\ast}$.)
One can also ask a similar question: given a representation $\rho$ of $g$ with $\ker\rho =0$ and highest weight $\Lambda$, how to compute the kernel of the resulting representation $R$ of the simply connected group $G$? The answer is as follows. The dual of the Lie algebra $t$ of $T$ contains the weight lattice $P$ and the root lattice $Q$. Recall that the weight lattice is the lattice spanned by the fundamental weights and the root lattice is spanned by the roots i.e. the weights of the adjoint representation. We have $Q\subset P$. Consider the dual lattices $Q^\vee\subset P^\vee$; these are formed by all elements $a$ of the Lie algebra of $T$ such that $l(a)\in\mathbf{Z}$ for all $l\in P$, resp. all $l\in Q$.
Inside $P^\vee$ there is a sublattice formed by all $a\in t$ such that $\Lambda(a)\in\mathbf{Z}$; it contains $Q^\vee$ and the quotient by $Q^\vee$ is naturally identified with the kernel of $R$ (by exponentiating). There is a similar result when $\rho$ is reducible -- in that case we just consider the sublattice formed by all elements of the tangent space such that all weights that take integral values on them.
Example: when $G=SL_2(\mathbf{C})$, we can identify $Q$ with the sublattice of $\mathbf{R}$ spanned by 1 and $R$ will be spanned by $\frac{1}{2}$. Then the representations with integral weight will have kernel $\mathbf{Z}/2$ and the representations with half-integral weight will have trivial kernel.
Hopefully one of these two questions also covers the questions of the posting.