Which is true and which false? I can't really decide which one is true and which false. Maybe in first 3 cases.
$$3n^5 − 16n + 2 \in O(n^5)$$ $$3n^5 − 16n + 2 \in O(n)$$ $$3n^5 − 16n + 2 \in \Omega(n^{17})$$ $$3n^5 − 16n + 2 \in \Omega(n^5)$$ $$3n^5 − 16n + 2 \in \Theta(n^5)$$ $$3n^5 − 16n + 2 \in \Theta(n)$$ $$3n^5 − 16n + 2 \in \Theta(n^{17})$$
and how to prove this one:
$$2^{(n+1)} \in O\left(\frac{3^n}n\right)$$
By definition, $f(n) = O(g(n))$ iff for $n$ big enough it holds that $|f(n)|\leq M |g(n)|$ for some positive constant $M$. That is, if you deal with polynomials, $$ P(n) = \sum_{k=0}^{m_1} a_k n^k,\quad Q(n) = \sum_{k=0}^{m_2} b_k n^k, \quad a_{m_1}, b_{m_2}>0 $$ the terms with the highest power dominate other terms and thus $$ P(n) = O(Q(n))\Leftrightarrow m_1\leq m_2 $$ which helps you resolving the first part of your question.
For the second part, apply similar analysis and show that for $n$ big enough it holds that $$ 2^{n+1} \leq \frac{3^n}{n} \text{ that is } \left(\frac{2}{3}\right)^n\leq \frac{1}{2n} $$
