Abelian variety with prescribed endomorphism ring

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 one may choose a rational number $c$ such that $End(P_{8,c})$ equals $End(P_{8,t})$ and therefore coincides with $\mathbb{Z}[\zeta_8]$.

The answer to your Question 1 is NO. See Th. 5.1 of my 2015 JPAA paper.

The answer to your Question 2 is YES. See 1963 Annals paper of Shimura "On analytic families ..".