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**NonCommAlg** there is nothing interesting about your second question. it is very straightforward. so i'll answer your first question for now and i'll help you with the second later if you tried and still couldn't do it.

for any lie algebra $\displaystyle L$ let $\displaystyle L^{(1)}=L^1=L' = [L,L]$ and inductively define $\displaystyle L^{(n+1)}=[L^{(n)},L^{(n)}]$ and $\displaystyle L^{n+1}=[L^n,L].$ an easy induction shows that $\displaystyle (L')^n \supseteq L^{(n+1)}$. thus if $\displaystyle L'$ is nilpotent, then $\displaystyle L$ is solvable.

conversely, suppose that $\displaystyle L$ is solvable. then, by Lie's theorem, $\displaystyle L$ has a basis $\displaystyle \mathfrak{B}$ such that $\displaystyle [\text{ad}(a)]_{\mathfrak{B}},$ the matrix of $\displaystyle \text{ad}(a)$ with respect to $\displaystyle \mathfrak{B},$ is upper triangular for every $\displaystyle a \in L.$ it is easy to see that if $\displaystyle A$ and $\displaystyle B$ are two upper triangular matrices, then $\displaystyle [A,B]=AB-BA$ is strictly upper triangular and hence nilpotent. thus $\displaystyle \text{ad}([a,b])=[\text{ad}(a), \text{ad}(b)]$ is nilpotent. so we have proved that $\displaystyle \text{ad}(c)$ is nilpotent for all $\displaystyle c \in L'$ and therefore, by Engel's theorem, $\displaystyle L'$ is nilpotent.