Let$~f~$be any function which is defined for all numbers. Show that$~g(x)=f(x)+f(-x)~$is even.
$$\operatorname{e.g.}~f(x)=x^2\implies g(x)=x^2+(-x)^2=2x^2\leftarrow~~\text{even}\tag{1}$$
$$\operatorname{e.g.}~f(x)=x^3\implies g(x)=x^3+(-x)^3=0\leftarrow~~\text{even}\tag{2}$$
$$\operatorname{e.g.}~f(x)=x^3+x^2~~\leftarrow~~\text{neither even nor odd}\tag{3}$$
$$\implies~g(x)=(x^3+x^2)+(-x^3+x^2)=2x^2\leftarrow\text{even}\tag{4}$$
But how it can be proven that the claim holds?
And needless to say, can I completely assume "for all numbers" in the problem statements belong to a set of complex numbers(handling imaginary numbers is also required in this problem)?
I do not think you define the concept of "even" or "odd" functions for functions of a complex variable. I will assume it is real. You have $$ g(-x)=f(-x)+f(-(-x))=f(-x)+f(x)=g(x) $$ so $g$ is even.