Let $Z=\mathbb{Z}$ be the ring of integers. This is a noetherian domain, and so satisfies property (a).

Let $A=\mathbb{Z}\langle x,y\ :\ x^2=2x,y^2=2y,xyx=2x,yxy=2y\rangle$. This is a (non-commutative) associative algebra with $1$. Further, it contains $Z$ as a central subalgebra. Moreover, $A$ is finitely generated as a $Z$-module, by the elements $\{1,x,y,xy,yx\}$. Thus condition (b) is satisfied.

Letting $K=\mathbb{Q}$, this is the field of fractions of $Z$, and $B=A\otimes_Z K\cong \mathbb{Q}\langle x,y\ :\ x^2=2x,y^2=2y,xyx=2x,yxy=2y\rangle$. We will write elements of $B$ using this isomorphism.

Since $A$ naturally embeds into $B$ (because $Z$ is regular in $A$) I'll take $A[b]$ to mean the subalgebra of $B$ generated by $A$ and $b$ (so we don't have to deal with Sasha's issue).

Consider the element $b_1=\frac{1}{2}x\in B$. It turns out that $A[b_1]$ is a finitely generated $Z$-module, generated by $\{1,\frac{1}{2}x,y,\frac{1}{2}xy,\frac{1}{2}yx\}$. Similarly, $A[b_2]$ is finitely generated, with $b_2=\frac{1}{2}y$.

However, $A[b_1,b_2]$ is not finitely generated, since alternating products $b_1b_2b_1b_2\cdots$ can give rise to arbitrarily large denominators. So the answer to (1) is no.

The algebra $A$ is fairly simple. We could even replace $\mathbb{Z}$ by the localization $\mathbb{Z}_{(2)}$, so $A$ would be a finitely generated module over a local ring.