 Solution

A nonempty weakly linear function has a domain $D$ closed under rational scalars and addition, hence a vector subspace of $\mathbb{R}$ over $\mathbb{Q}$. Additivity first gives $f(0)=0$ and $f(-x)=-f(x)$, then $f(nx)=nf(x)$ for integers $n$. If $q=m/n$ with $n>0$, rational closure puts $qx$ in $D$ andThus $f(qx)=qf(x)$: it is a $\mathbb{Q}$-linear map. The empty function is not maximal because it extends to the function on $\{0\}$ with value zero.

Suppose a maximal $f$ had proper domain $D\ne \mathbb{R}$. Choose $t\notin D$ and any prescribed real value $a$, for instance $a=0$. Every element of $D+\mathbb{Q}t$ has a unique expression $d+qt$: if two expressions coincide and the coefficients differ, then $t$ would belong to $D$. DefineUniqueness makes this well-defined. The displayed logical formula and the $\mathbb{Q}$-linearity of $f$ show directly that $h$ is additive and has a domain closed under rational scalars. Hence it is a weakly linear function and extends $f$. It is a proper extension because $t$ is now in its domain, contradicting maximality. ThereforePart (a) and Zorn's lemma guarantee the existence of maximal extensions as well as this description of their domains.