Main question: For any given sequence $\mathfrak{g}_1\xrightarrow{f_1}\mathfrak{g}_2\xrightarrow{f_2}\mathfrak{g}_3$ of finite dimensional Lie algebras with $\mathrm{Im}\,f_1 = \mathrm{Ker}\,f_2$, is there a sequence $G_1\xrightarrow{F_1}G_2\xrightarrow{F_2}G_3$ of Lie groups satisfying $\mathrm{Lie}(G_i) = \mathfrak{g}_i$, $dF_i = f_i$ and $\mathrm{Im}\,F_1 = \mathrm{Ker}\,F_2$?
My attempt was as following: There exist 1-connected $G_i$'s corresponding to $\mathfrak{g}_i$'s and also $F_i$'s corresponding to $f_i$ thanks to Lie group-Lie algebra correspondence. Since $f_2\circ f_1=0$, $F_2\circ F_1$ is trivial; hence $\mathrm{Im}\,F_1 \subset \mathrm{Ker}\,F_2$. While the inverse inclusion is not obvious, in the case that $\mathrm{Ker}\,F_2$ is connected, $\mathrm{Im}\,F_1 = \mathrm{Ker}\,F_2$ holds. Is it always the case? Namely,
Additional question: Let $G,H$ be 1-connected Lie groups and $f\colon G\to H$ be any Lie group homomorphism. Is $\mathrm{Ker}\, f$ always connected?
Thanks!
The answer to your additional question is negative. Consider the following example:
We define a map $\mathbb{R}\to \mathrm{SL}_2(\mathbb{C})$ by $$\alpha\mapsto \begin{bmatrix}\exp(i\alpha)&0 \\ 0 & \exp(-i\alpha)\end{bmatrix}$$ This map has kernel $2\pi\mathbb{Z}$. More generally if $G$ is a 1 connected Lie group with nontrivial maximal torus $T$, any one paremeter subgroup of the torus defines a homomorphism $\phi: \mathbb{R}\to G$ for which $\ker\phi\cong \mathbb{Z}$.