Answer
a. See explanations.
b. The function will not be negative.
Work Step by Step
a. With $f(x)=3+4cos(x)+cos(2x)$, the first term is a constant, the second term $4cos(x)$ has a period of $p_1=2\pi$, the third term $cos(2x)$ has a period of $p_2=\pi$, thus the overall period of the function is the larger one $p=2\pi$ which means we only need to consider an interval of $[0,2\pi]$ (see graph).
b. Take the derivative to get
$f'(x)=-4sin(x)-2sin(2x)=-4sin(x)-4sin(x)cos(x)=-4sin(x)(1+cos(x))$. Let $f'(x)=0$, we get $x=0,\pi$ in $[0,2\pi]$, and $f(0)=3+1+1=5\gt0$ and $f(\pi)=3-4+1=0$.
Testing sign changes of $f'(x)$ across $x=0,\pi$, we get
$..(+)..(0)..(-)..(\pi)..(+)..$. Thus $f(0)=5$ is a maximum and $f(\pi)=0$ is a minimum, indicating $f(x)\geq0$ for all $x$ over the domain. The function will not be negative.
