Back-quotes(``) and $() can be both used for command-substitution, but they have minor difference.
Take the cases mentioned in the question as examples:
$echo `echo \\\\ `
The 1st and 3rd "\" will be treated as escapes, echo \\\\ will be evaluated as "\\"
Therefore, the above command is equal as:
$echo \\
And the first backslash is treated as escape, so the output is:
\
In the case of $(), there is a little tricky, what is evaluated inside $() will be passed as an argument to the outside command.
As an example:
$echo $(echo \\\\ )
What is inside the $() is evaluated as "\\", which is the same as the previous case. What's different is that "\\" will be passed directly to the outside echo command, the first backslash will not be treated as an escape.
Therefore we will got the output:
\\
