Don't use the equals sign! $o(n \log_2 n)$ is the set of functions whose growth rate is strictly less than $n \log_2\,n$. So the proper expression is $f(n) \in o(n \log_2 n)$. Similarly, $O(n \log_2 n)$ is the set of functions whose growth rate is less than or equal to $n \log_2\,n$.
The simple way to determine which expressions are true is to look at the fastest growing terms. In $f(n)$ that is $4n^2$ which clearly grows faster than $2n + 1$ but slower than $n^4$.
I suggest using Python to find out for yourself:
>>> from math import *
>>> def f(n): return 4*n*(n + 2*log(n**2)**2) + e**(-n) + 8*sin(2*pi*n/256)
>>> n = 999999999
>>> f(n) < 2**n
So $f(n) \in o(2^n)$ is likely true.
>>> f(n) < 2*n + 1
So $f(n) \in O(2n + 1)$ is false. While this is not a rigorous method it allows you to get a feel for the numbers. See also the answer here https://stackoverflow.com/questions/1364444/difference-between-big-o-and-little-o-notation