Are there numbers $k > 1$ and $c > 1$ such that:

1 ) $\theta(c) \geq c \left( 1-\frac{1}{5 \ln^2(c)} \right) $

2 ) $\frac{c}{1+\frac{1}{\ln^4(c)}} \leq p(\pi(c))$ where $p(n)$ is the $n$-th prime number and $\pi(n)$ is the prime counting function

3 ) $\pi(c) \leq \frac{c}{\ln (c)} (1+\frac{1}{\ln (c)}+\frac{2}{\ln^2 (c)}+\frac{6}{\ln^3 (c)} + \frac{25}{\ln^4 (c)} )$

4 ) $\pi(c) \geq \frac{c}{\ln (c)} (1+\frac{1}{\ln (c)}+\frac{2}{\ln^2 (c)}+\frac{6}{\ln^3 (c)} + \frac{24}{\ln^4 (c)} )$

5 ) $\prod \limits_ {p \leq c} (1+\frac{1}{p^2-1}) \prod \limits_{p \leq c^k} (1+\frac{1}{p}) \leq e^\gamma \ln (c ( 1-\frac{1}{5 \ln^2(c)})+c^k ( 1-\frac{1}{5 k^2 \ln^2(c)}))$?