An abelian group $D$ is said to be divisible if given any $y \in D $ and $n\neq0 \in Z$, there exists $x \in D$ such that $nx = y$.
(a) No nonzero finite abelian group is divisible.
(b) No nonzero free abelian group is divisible.
(c) $\mathbb{Q}$ is a divisible abelian group.
I have done (c). For (a) and (b) I tried by assuming in both the cases that there exists an element $a$ such that $a= n b$ for some $n\in \mathbb{Z}\neq 0$ and $b$ in group in hope of getting some contradiction:
(a) $a=nb$ implies that $a b^{-1} =n$ which means that n also belongs to group which is absurd, contradiction and similarly contradiction will be obtained for (c).
Is my solution correct? How to approach solution when $n$ belongs to the group $D$?
In case I am wrong please tell the correct solution.
No, this is wrong. The "$nb$" notation means adding $b$ to itself $n$-times. It is not really a multiplication. We use this multiplicative notation only because it coincides with a multiplication of numbers. But we are not dealing with numbers here, we are dealing with an arbitrary abelian group with addition. And so neither $b^{-1}$ nor multiplying by it is defined.
Of course every finite abelian group can be extended into a unitary ring and but then (1) you have to explain why, probably by invoking classification of finite abelian groups, (2) $b$ does not have to be invertible, (3) in such situation you obtain $ab^{-1}=n\cdot 1$ where $1$ is the multiplicative identity (so still no contradiction) and (4) this an overcomplicated approach to the problem.
And it is not clear to me whether every free abelian group can be turned into a unital ring. They can if they are of finite rank but for infinite rank? That's an interesting question, I don't know to be honest. And so this is similar to previous case, except point (1) is harder (if even true).
For (a) note that if $G$ is finite of order $m$ then $mx=0$ for any $x\in G$. And so if $G$ has at least two elements, one of which is nonzero, say $y\neq 0$ then $mx=y$ equation has no solution.
For (b) note that if $G$ is free abelian then $G$ is isomorphic to a direct sum $\bigoplus_{i\in I}\mathbb{Z}$. Define $(a_i)\in \bigoplus_{i\in I}\mathbb{Z}$ by $a_i=1$ for any $i$. Then the equation $2(x_i)=(a_i)$ has no solution, because $2x=1$ has no solution in $\mathbb{Z}$.