Solutions

i) We have $f(-2)=-7$ and $f(-1)=1$. There exists an $x\in[-2,-1]$ such that $f(x)=0$ by the Intermediate Value Theorem (IVT). Hence, $n=-2$.

ii) We have $f(-5)=-9$ and $f(-4)=4^4-7$. Hence, $n=-5$ by the IVT.

iii) We have $f(-1)=-1$ and $f(0)=1$. Hence, $n=-1$ by the IVT.

iv) Notice that

$$f(x)=4x^2-4x+1=(2x-1)^2.$$

Its root is $x=\frac12$. Hence, $n=0$ by the IVT.