**Summary** The answer to the first question is affirmative and to the second question is negative, but for rather mundane reasons. In the simple Lie algebra case, cyclicity of ${\rm ad}\, a$ for some $a$ implies that $\frak{g}$ has rank $1$, in which case every non-zero element $x\in\frak{g}$ is cyclic in every simple module. On the other hand, for any $\frak{g}$, every $x\in\frak{g}$ acts cyclically on any one-dimensional module (e.g. the trivial representation). Thus in the higher rank case, all elements (even $x=0$) are trivial-cyclic and none are ad-cyclic. Moreover, all modules $V$ which admit a semisimple cyclic $x\in\frak{g}$ are weight multiplicity-free (WMF), which have been classified, and the corresponding cyclic elements can be explicitly described.

**Details** Assume that the ground field is not of characteristic $2$. It is easy to see that a simple Lie algebra $\frak{g}$ contains an ad-cyclic element only if $\frak{g}$ has rank $\ell=1$, so that ${\frak g}$ is a form of ${\frak sl}_2$. Indeed, by the standard properties of endomorphisms, the condtion "$A$ is cyclic" implies that the dimension of $\ker{A}$ is at most one, whereas for $A={\rm ad}\, x$, this dimension is at least $\ell$ by the definition of rank; thus $\ell\leq 1$. Conversely, every non-zero element $x$ of a simple rank $1$ Lie algebra acts cyclically in every finite-dimensional simple module: extend the scalars to an algebraically closed field and consider separately the cases of semisimple and nilpotent $x$.

A **semisimple** endomorphism (or a diagonal matrix) is cyclic iff its eigenvalues are distinct. Let $h\in{\frak h}$ be a semisimple element of a split semisimple Lie algebra $\frak{g}$ with a Cartan subalgebra ${\frak h}$ and $V$ an $N$-dimensional $\frak{g}$-module with weights $\lambda_1,\ldots,\lambda_N\in {\frak h}^*$. Then the matrix of $h$ relative to a weight basis of $V$ is diagonal with eigenvalues $\lambda_i(h)$ and $h$ acts cyclically on $V$ iff $\lambda_i(h)$ are pairwise non-equal. In particular, this is possible only if $V$ is weight multiplicity-free (WMF), and all such modules have been classified (consistent with the above, the adjoint module is WMF iff ${\frak g}$ is a direct sum of several copies of ${\frak sl}_2$). For example, in characteristic $0$, if ${\frak g}={\frak sl}_n$, the simple WMF modules are exhausted by the symmetric powers $S^k W$ of the defining module $W$, their duals, and the exterior powers $\Lambda^k W$ of $W$, and the necessary condition for $h$ to act cyclically on $V$ is that $h$ is regular (equivalently, $h$ acts cyclically on $W$, i.e. has distinct eigenvalues). Of course, this condition is not sufficient in general. For example, if $V=S^2 W$ and $h$ is diagonal with eigenvalues $a_i$ that sum to $0$, $h$ is cyclic on $V$ iff all $a_i+a_j, 1\leq i\leq j\leq n$ are pairwise distinct. For $n\geq 3$, this is clearly stronger than just all $a_i$ being pairwise distinct.

Note, however, that reducible WMF modules also exist: for example, let $\frak{g}={\frak sl}_n$ and $V=W\oplus\Lambda^2 W$, then $V$ is WMF and every diagonal $h$ with $n$ distinct eigenvalues $a_i$ satisfying appropriate additional inequalities acts cyclically on $V$. Thus contrary to OP's Remark 1, an element of ${\frak g}$ *can* act cyclically on a reducible module.

If $e$ is a **nilpotent** element acting cyclically on a $\frak{g}$-module $V$ then $V$ has to be simple with respect to the corresponding ${\frak sl}_2$-subalgebra, so that the weights of $V$ belong to a single line. This places severe constraints on the possible data. For a simple classical Lie algebra ${\frak g}\ne{\frak sl}_2$ and a non-trivial module $V$, this leaves only the cases of ${\frak g}={\frak sl}_n$ and $V$ the defining module or its dual, and a symplectic or odd orthogonal ${\frak g}$ and the defining ("vector") module $V$, with the element $e$ being a principal nilpotent in all cases.

Finally, if $x=s+n$ is the Jordan decomposition of a **general** element acting cyclically on a $\frak{g}$-module $V$ then the centralizer $Z(s)$ of the semisimple part $s$ is a reductive subalebra containing $x$, so that $V$ is a $Z(s)$-module with a cyclic action of $x$. This provides an approach to classifying general cyclic pairs $(V,x)$.

vectors(which I think led to the confusion with @JimHumphreys's answer), whereas your title suggests that you are interested in cyclicelements(those that have a cyclic vector in the adjoint representation). $\endgroup$ – LSpice Jul 4 '19 at 14:05