= Solution
The <Cauchy functional equation> first gives $g(0)=0$ and $g(-x)=-g(x)$. If $h\to0$, then continuity at the given point $z$ gives
$$
g(h)=g(z+h)-g(z)\longrightarrow0,
$$
so $g$ is continuous at zero. For every $x$,
$$
g(x+h)-g(x)=g(h)\longrightarrow0,
$$
so $g$ is continuous everywhere.
Put $c=g(1)$. Additivity gives $g(n)=nc$ for integers $n$ and then $g(m/n)=(m/n)c$ for <rational numbers>. For any real $x$, choose rationals $q_j\to x$; continuity gives
$$
\boxed{g(x)=\lim_jg(q_j)=\lim_jcq_j=cx}.
$$
For the multiplicative equation, $h(0)=h(0)^2$. If $h(0)=0$, then $h(x)=h(x)h(0)=0$ for every $x$. Otherwise $h(0)=1$. In that case
$$
h(x)=h(x/2)^2\geq0,
$$
and $h(x)$ cannot vanish because $1=h(0)=h(x)h(-x)$. Hence $h$ is everywhere positive. The continuous function $g=\log h$ satisfies the <Cauchy functional equation>, so $g(x)=cx$. Therefore the complete family is
$$
\boxed{h\equiv0\quad\text{or}\quad h(x)=e^{cx}\ \text{for some }c\in\mathbb R}.
$$
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