To summarize the material in the comments so that the question does not appear as unanswered...

(1) The first question has a negative answer as stated. For example, take $$G = \langle x,y\mid x^8=y^8=[x,y]^2=[x,y,x]=[x,y,y]=1\rangle$$ a $2$-group of order 128 and class $2$. Take $M=\langle [x,y],x\rangle$, so that $G/M$ is of order $8$. Since $Z(G)=\langle x^2,y^2,[x,y]\rangle$, then $MZ(G) = \langle x^2,y,[x,y]\rangle\neq G = C_G(G')$.

Question (2) seems a bit harder to answer precisely.

For Question (3), Derek Holt provides the following:

Proposition. Let $G$ be a metabelian $p$-group. The following are equivalent:

1. $C_G(G')=G'$.
2. $G'$ is maximal among abelian subgroups of $G$.
3. $G'$ is maximal among abelian normal subgroups of $G$.

Proof. (1)$\implies$(2) If $G'\subseteq A$ with $A$ abelian then $A\subseteq C_G(G')$. Hence $G'=A$.

(2)$\implies$(3) Immediate.

(3)$\implies$(1) That $G'\subseteq C_G(G')$ follows because $G$ is metabelian. And any subgroup of $G$ that contains $G'$ is normal; so if $x\in C_G(G')$, then $\langle G',x\rangle$ is abelian and normal, so by maximality $x\in G'$. $\Box$

To summarize the material in the comments so that the question does not appear as unanswered...

(1) The first question has a negative answer as stated. For example, take $$G = \langle x,y\mid x^8=y^8=[x,y]^2=[x,y,x]=[x,y,y]=1\rangle$$ a $2$-group of order 128 and class $2$. Take $M=\langle [x,y],x\rangle$, so that $G/M$ is of order $8$. Since $Z(G)=\langle x^2,y^2,[x,y]\rangle$, then $MZ(G) = \langle x^2,y,[x,y]\rangle\neq G = C_G(G')$.

Question (2) seems a bit harder to answer precisely.

For Question (3), Derek Holt provides the following:

Proposition. Let $G$ be a metabelian $p$-group. The following are equivalent:

1. $C_G(G')=G'$.
2. $G'$ is maximal among abelian subgroups of $G$.
3. $G'$ is maximal among abelian normal subgroups of $G$.

Proof. (1)$\implies$(2) If $G'\subseteq A$ with $A$ abelian then $A\subseteq C_G(G')$. Hence $G'=A$.

(2)$\implies$(3) Immediate.

(3)$\implies$(1) That $G'\subseteq C_G(G')$ follows because $G$ is metabelian. And any subgroup of $G$ that contains $G'$ is normal; so if $x\in C_G(G')$, then $\langle G',x\rangle$ is abelian and normal, so by maximality $x\in G'$. $\Box$