Let $G,H$ be two compact abelian groups, let $\mathbb{T}$ be the torus (also known as $\mathbb{R}/\mathbb{Z}$ or $S^1$).
I have the following exact sequence $$ 1\rightarrow \mathbb{T} \rightarrow G \rightarrow H \rightarrow 1 $$ Can I conclude that $ G=\mathbb{T}\oplus H$?
Yes: The category of compact abelian groups is equivalent to the opposite of the category of abelian groups via the character functor $G\mapsto \widehat{G}:=\text{Hom}(G,T)$ (see Pontryagin duality), and $T$ corresponds to $\widehat{T}\cong{\mathbb Z}$ under this equivalence. As the latter is projective in the category of abelian groups, the dual exact sequence $$0\to \widehat{H}\to\widehat{G}\to\widehat{T}\to 0$$ splits. In particular, we have $\widehat{G}\cong\widehat{T}\oplus\widehat{H}\cong\widehat{T\oplus H}$, hence $G\cong T\oplus H$, too.