A homomorphism from $(\mathbb{Q},+)\rightarrow (\mathbb{Q},+)$

I claim that any group homomorphism will be of the form $f(x)=qx$ for some $q\in\mathbb{Q}$, now as $f$ is non zero so $q\neq 0$ now clearly kernel of $f$ will be $\{0\}$ only hence $f$ is injective, now let $f(1)=p\neq 0$ then for any $y\in\mathbb{Q}$ and $f(y/p)=y$ so $f$ is surjective too, hence bijective. am I right?

You are right. To see that all group homomorphisms have the form you claim, is not completely trivla, as $\mathbb Q$ is not even finitely generated. But note that for $n\in\mathbb Z$ you have $f(n\cdot x)=n\cdot f(x)$, hence for $q=\frac ab$ with $b\ne 0$ it follows indeed that $$f(q)=\frac1b\cdot f(b\cdot q)=\frac1b\cdot f(a)=\frac1b\cdot a\cdot f(1)=q\cdot f(1).$$ From this your observations about a,b,c follow just as you write.