a) Let's find a third number to the two numbers under (1) so that the sum of the three numbers is equal to their product.

b) Can a two-variable function be given whose values at appropriate points are equal to these three numbers, respectively?

$$
\frac{a+b}{1-a b}, \quad \frac{b+c}{b c-1}
$$
a) Assuming that the numbers under (1) - let's denote them by $E$ and $M$ - exist, that is,

$$
a b \neq 1, \quad b c \neq 1
$$

Then the requirement for the sought third number $H$ is:

$$
E+M+H=E M H
$$

From this,

$$
H=\frac{E+M}{E M-1}
$$

or by substitution,

$$
H=\frac{\frac{a+b}{1-a b}+\frac{b+c}{b c-1}}{\frac{(a+b)(b+c)}{(1-a b)(b c-1)}-1}=\frac{(a+b)(b c-1)+(b+c)(1-a b)}{(a+b)(b+c)-(1-a b)(b c-1)}=\frac{c-a+b^{2}(c-a)}{1+a c+b^{2}+a b^{2} c}=\frac{(c-a)\left(1+b^{2}\right)}{(1+a c)\left(1+b^{2}\right)}
$$

from which, by simplification,

$$
H=\frac{c-a}{1+a c}
$$

since $1+b^{2} \geq 1$, so it is a non-zero number. Thus, $H$ exists and meets the requirement, provided that $a c \neq-1$.
b) The denominators of $E$ and $H$ are both obtained by subtracting the product of the two terms of the numerator from 1, and the same holds for the second term as well according to the transformation

$$
M=\frac{(-b)+(-c)}{1-(-b)(-c)}
$$

We can say: our numbers are equal to the values of the function of the variables $x, y$

$$
\frac{x+y}{1-x y}
$$

in the cases

$$
x=a, y=b ; \quad x=-b, y=-c ; \quad x=c, y=-a
$$

According to (2), $H$ is equal to the value of the same function at the point $x=-E, y=-M$.