How do you call a function defined in .bashrc from the shell?

Any changes made to .bashrc will only take effect in a new terminal session. If you want to apply the changes to your current terminal session, you have to instruct the shell to re-read the .bashrc. The shortest way to to this is to use the . command, which is a synonym to source:

[user@linuxPc]$ . ~/.bashrc

You can export functions. In your ~/.bashrc file after you define the function, add export -f functionname.

function hello() {
   echo "Hello, $1!"
}

export -f hello

Then the function will be available at the shell prompt and also in other scripts that you call from there.

Note that it's not necessary to export functions unless they are going to be used in child processes (the "also" in the previous sentence). Usually, even then, it's better to source the function into the file in which it will be used.

Edit:

Brackets in Bash conditional statements are not brackets, they're commands. They have to have spaces around them. If you want to group conditions, use parentheses. Here's your function:

function coolness() {

    if [ -z "$1" -o -z "$2" ]; then
        echo "Usage: $0 [sub_package] [endpoint]";
        exit 1;
    fi
        echo "Hi!"
}

A better way to write that conditional is:

    if [[ -z "$1" || -z "$2" ]]; then

because the double brackets provide more capability than the single ones.

The test in your function won't work - you should not have brackets around the -z clauses, and there should be a space between if and the open bracket. It should read:

function coolness() {

    if [ -z "$1" -o -z "$2" ]; then
        echo "Usage: $0 [sub_package] [endpoint]";
        exit 1;
    fi
    echo "Hi!"
}

Include in your script the line

source .bashrc

try with the source construct it should work!