This is a statement made in Knapp, Lie groups, Lie algebras, Cohomology Chpt 4 last paragraph of Sec 2.
$H^i(g,a)$ is the $i-$th cohomology group of complex $Hom(\wedge^i g,a)$ with $a$ abelian lie algebra.
"If $g$ is semisimple, it is not too hard to see that $H^2(g,a)=0$. With a little supplementary argument, it follows that any finite-dimensional complex lie algebra is semidirect product of a semisimple lie algebra and a solvable lie algrbra. The same theorem and proof apply over $R$ and a structure theorem for lie groups drops out. Results for $H^1$ then furnishes a uniqueness theorem"
$\textbf{Q1:}$ Maybe this is too obvious for others. I do not see why obviously $H^2(g,a)=0$. Fix $\pi:g\to End(a)$ representation. I am using Brian, Hall's Lie algebra's semisimple meaning reductive and trivial center. I could see all semi simples are decomposed into simples and this is unique upto reordering. So given an extension $0\to a\to h\to g\to 0$. Now $h=a\oplus_\pi g$ as semi-direct product but the lie bracket between $a,g$ will be twisted by representation of $g\to End(a)$. I do not see why all extensions are equivalent to $0\to a\to a\oplus_\pi g\to g$.
$\textbf{Q2:}$ Why it follows that any finite-dimensional complex lie algebra is semidirect product of a semisimple lie algebra and a solvable lie algrbra?
$\textbf{Q3:}$ Why the same theorem and proof apply over $R$ and a structure theorem for lie groups drops out? Over $R$, I no longer have unitary representation which may not allow me to have decomposition of $g$ into simple ones.
$\textbf{Q4:}$ What are results for $H^1$ furnishing a uniqueness theorem? What is the meaning of uniqueness theorem here?