The answer to your Question 3 is YES with the ground field $\mathbb{Q}$.

Here is a sketch of the proof. For each positive integer $q$ and a "parameter" $t$ (in char 0) consider the smooth projective model $C_{q,t}$ of an affine curve $y^q=x^3-x-t$. Let $P_{8,t}$ be the Prym variety of the double cover
$$C_{8,t}\to C_{4,t}, (x,y)\mapsto (x,y^2).$$
Then $P_{8,t}$ is an abelian fourfold provided with an embedding 
$$\mathbb{Z}[\zeta_8]\hookrightarrow End(P_{8,t}).$$
One may deduce from Theorem 1.5 of  arXiv:math/0601072 [math.AG] that if $t$ is  a transcendental number then $P_{8,t}$ does not contain positive-dimensional abelian subvarieties of CM type. It follows from  Th. 1.1 of a paper by Jiangwei Xue and Chia-Fu Yu (arXiv:1304.6202 [math.NT]) that for such $t$ the endomorphism ring $End(P_{8,t})$ coincides with $\mathbb{Z}[\zeta_8]$. Now, by Masser's specialization theorem http://archive.numdam.org/ARCHIVE/BSMF/BSMF_1996__124_3/BSMF_1996__124_3_457_0/BSMF_1996__124_3_457_0.pdf,
one may choose a rational number $c$ such that
$End(P_{8,c})$ equals $End(P_{8,t})$
and therefore coincides with  $\mathbb{Z}[\zeta_8]$.