a) Consider $A:={\bf Z},B:={\bf Q}$ as additive groups and define the homomorphism $f:A\rightarrow C:={\bf Q}^*$ by $f(k)=2^k$. If it has an extension homomorphism $F:B\rightarrow C$ then $F(1/2)^2=F(1)=f(1)=2$ but $x^2=2$ has no solution in $C$.

b) Let $H$ be an abelian group, and $H'\leq H$ be a subgroup with a homomorphism $\mu'\colon H\to(\mathbb{Q},\,+)$. Consider the subgroup $F:=\{g\in H\mid\mathrm{ord}(g)<\infty\}$ of elements of finite order. It is indeed a group: for $g,h\in F$, we have \[ \mathrm{ord}(g^{-1})=\mathrm{ord}(g)<\infty\implies g^{-1}\in F \]and \[ (gh)^{\mathrm{ord}(g)\mathrm{ord}(h)}=(g^{\mathrm{ord}(g)})^{\mathrm{ord}(h)}(h^{\mathrm{ord}(h)})^{\mathrm{ord}(g)}=e\implies\mathrm{ord}(gh)\leq\mathrm{ord}(g)\mathrm{ord}(h)<\infty\implies gh\in F. \]Since $H$ is abelian, $F\trianglelefteq H$ so let $A:=H/F$ and $A':=H'/(F\cap H')$. Then $A$ is an abelian group with no nontrivial elements of finite order: if $a^d=e_A$ for $a\in A\setminus\{e_A\}$ and $d\in\mathbb{Z}_{\geq 2}$, we can write $a=:hF$ so \[ h^d\in F\implies\mathrm{ord}(h)\leq d\cdot\mathrm{ord}(h^d)<\infty\implies h\in F\implies a=e_A, \]a contradiction. Since every nonzero element of $(\mathbb{Q},\,+)$ has infinite order, $\mu'(F\cap H')=0$. Set $\mu(F)=0$, which agrees with $\mu'$ on $F\cap H$. Then $\mu$ must be constant on each coset of $F$ in $H$. Thus $\mu'$ induces a homomorphism $\varphi'\colon A'\to(\mathbb{Q},\,+)$ by $\varphi'(hF):=\mu'(h)$. Any extension $\varphi\colon A\to(\mathbb{Q},\,+)$ of $\varphi'$ induces an extension $\mu\colon H\to(\mathbb{Q},\,+)$ of $\mu'$ by $\mu(h):=\varphi(hF)$. Thus, it suffices to extend $\varphi'$ to a homomorphism $\varphi\colon A\to(\mathbb{Q},\,+)$. From now on, $e$ will denote $e_A$. Lemma. For every $a\in A$ and $d\in\mathbb{Z}^+$, there exists at most one $b\in A$ for which $b^d=a$. Proof. If $b_1^d=a=b_2^d$, we have $(b_1b_2^{-1})^d=e$, so $b_1=b_2$ since $A$ has no nontrivial elements of finite order. $\square$ Hence, if $b^n=a$ for $n\in\mathbb{Z}^+$, we can write $b=a^{\frac{1}{n}}$ as $a^{\frac{1}{n}}$ is unique. Let $B$ be the group consisting of the unique $n$th roots of the elements in $A$ (if the $n$th root of $a$ is not in $A$, create a new element for it): \[ B:=\{a^{\frac{1}{n}}\mid a\in A,\,n\in\mathbb{Z}^+\}. \]$B$ is indeed a group: we have $(a^{\frac{1}{n}})^{-1}=(a^{-1})^{\frac{1}{n}}$ and $a_1^{1/n_1}a_2^{1/n_2}=(a_1^{n_2}a_2^{n_1})^{\frac{1}{n_1n_2}}$. Note that $A$ is a subgroup of $B$. Additionally, $B$ is a $\mathbb{Q}$-vector space with scalar multiplication defined as $c\cdot b:=b^c$. Take a basis $\mathcal{B}'\subseteq A'$ for $\mathrm{span}(A')$ and extend it to a basis $\mathcal{B}$ for $B=\mathrm{span}(A)$. Let homomorphism $\varphi\colon A\to(\mathbb{Q},\,+)$ be generated by $x\mapsto \varphi'(x)$ for $x\in\mathcal{B}'$, and $x\mapsto 0$ for $x\in\mathcal{B}\setminus\mathcal{B}'$. Any $a\in A'$ can be written as $a=\prod_{x\in S}x^{q_x}$ for a finite subset $S\subseteq\mathcal{B}'$ and rationals $\{q_x\mid x\in S\}\subseteq\mathbb{Q}$. There exists an integer $n\in\mathbb{Z}$ such that $nq_x$ is an integer for all $x\in S$ (taking the product of the denominators of the $q_x$ works). Then \[ \varphi(a)=\sum_{x\in S}q_x\varphi(x)=\frac{1}{n}\sum_{x\in S}nq_x\varphi'(x)=\frac{1}{n}\varphi'\left(\prod_{x\in S}x^{nq_x}\right)=\frac{1}{n}\varphi'(a^n)=\varphi'(a), \]so $\varphi$ agrees with $\varphi'$ on $A'$. Thus $\varphi$ is our desired extension homomorphism. $\square$

An abelian group with "property (P)" is called "divisible group" in the mathematical literature : see the Wikipedia page "Divisible groups".