Hi
For the first one, you can try to show that $\displaystyle \left|\frac{n^3+(-1)^n8n^2+1}{n^4+(-1)^{n+1}n^2}\right|\sim\frac{1}{n}$ for $\displaystyle n\to\infty$. For the second one, the ratio test may be helpful.
Not completely off-topic question : If $\displaystyle f \sim g$ do English speaking people say that $\displaystyle g$ is an approximation of $\displaystyle f$ ? Is there another way of saying this ?
You can get it by showing that
$\displaystyle \frac{\left|\frac{n^3+(-1)^n8n^2+1}{n^4+(-1)^{n+1}n^2}\right|}{\frac{1}{n}}=\left|\frac{n^4+ (-1)^n8n^3+n}{n^4+(-1)^{n+1}n^2}\right|$
tends to 1 as $\displaystyle n$ approaches $\displaystyle \infty$. (divide the numerator and the denominator by $\displaystyle n^4$)