, 1997

The only solution is $f(x) = x$ for any $x \in \mathbb{R}$.

Using $f(x+1) = f(x) + 1$, it can be proved easily by induction that
$$f(x+n) = f(x) + n \quad (1)$$
for any integer $n$.

Consider any rational number $\frac{a}{b}$ where $a, b \in \mathbb{Z}$. For any positive integer $k$, we substitute $x = \frac{a}{b} + kb^2$ in (ii) to obtain
$$f\left(\frac{a^3}{b^3} + 3ka^2 + 3k^2ab^3 + k^3b^6\right) = f(x^3) = f(x)^3.$$
Since $k, a, b \in \mathbb{Z}$, by (1), we obtain
$$f\left(\frac{a^3}{b^3} + 3ka^2 + 3k^2ab^3 + k^3b^6\right) = f\left(\frac{a^3}{b^3}\right) + 3ka^2 + 3k^2ab^3 + k^3b^6.$$
Also, by letting $z = f\left(\frac{a}{b}\right)$, we have
$$f(x)^3 = (z + kb^2)^3 = z^3 + 3kb^2z^2 + 3k^2b^4z + k^3b^6.$$
Comparing the two sides, since $f\left(\frac{a^3}{b^3}\right) = z^3$, we find that
$$3ka^2 + 3k^2ab^3 = 3kb^2z^2 + 3k^2b^4z.$$
This implies $b^2z^2 + kb^4z - a^2 - kab^3 = 0$, and hence
$$(bz - a)(bz + a + kb^3) = 0.$$
Since $b \neq 0$, we cannot have $bz + a + kb^3 = 0$ for all $k \in \mathbb{Z}^+$. Therefore, we must have $bz - a = 0$, i.e. $z = \frac{a}{b}$. This proves $f(x) = x$ for any $x \in \mathbb{Q}$.

Since $\mathbb{Q}$ is dense in $\mathbb{R}$, for any $x \in \mathbb{R}$, we can find a sequence $\{x_n\}$ of rational numbers converging to $x$. By the continuity, we have
$$f(x) = f\left(\lim_{n \to \infty} x_n\right) = \lim_{n \to \infty} f(x_n) = \lim_{n \to \infty} x_n = x.$$
It is easy to check that the identity function satisfies both conditions, so it is the only solution.