= Solution
For the <additive function>, setting both arguments to zero gives $f(0)=0$, and then $f(-x)=-f(x)$. Continuity at the given point $x_0$ implies
$$
f(h)=f(x_0+h)-f(x_0)\longrightarrow0\qquad(h\to0).
$$
For any real $x$, $f(x+h)-f(x)=f(h)$, so $f$ is continuous at every point. This proves the continuity part of <additive function continuous at one point is linear>.
Let $c=f(1)$. Repeated addition and negation give $f(m)=mc$ for every integer $m$. For integers $p$ and positive $q$,
$$
qf(p/q)=f(p)=pc,
$$
so $f(r)=cr$ for every <rational number> $r$. Given any real $x$, choose rational $r_n\to x$. Continuity and <density of the rational numbers> give
$$
\boxed{f(x)=\lim_n f(r_n)=\lim_n cr_n=cx.}
$$
Conversely each real multiple of $x$ satisfies the <Cauchy functional equation>.
For the multiplicative equation, assume first that $g$ is not identically zero and choose $x_1$ with $g(x_1)\ne0$. Then $g(x_1)=g(x_1)g(0)$ forces $g(0)=1$. Next
$$
g(x)g(-x)=1
$$
makes every value nonzero, and
$$
g(x)=g(x/2)^2>0
$$
makes every value positive. Thus $h(x)=\log g(x)$ is a continuous real <additive function>, since $h(x+y)=h(x)+h(y)$. The proved classification gives $h(x)=kx$ for a real constant $k$. This solves the <continuous exponential functional equation>:
$$
\boxed{g(x)\equiv0\quad\text{or}\quad g(x)=e^{kx}\ (k\in\mathbb R).}
$$
Both types satisfy the equation and continuity, and every nonzero solution is everywhere strictly positive.
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