About Riemann's Hypothesis. Could Riemann' Hypothesis be proven true using Robin's Inequality and that a counter-example to Riemann's Hypothesis can not have a divisor that is a prime number to the exponent 5 ,according to some of Robin's Theories? Also I think it can be proven the product of two numbers A and B that are counter-examples to R.H. is also a counter-example. Note a number $n$ is a counter-example to R.H. if $\sigma(n)\gt e^{\gamma}n\ln\ln n$ (forgive notation).
 A: Robin's inequality was proved to be true if the Riemann Hypothesis holds, so disproving Robin's inequality would be one way of disproving the Riemann Hypothesis. However, the inequality has been around for a long time (Ramanujan got the result in 1915), this is probably just as hard as any other way of settling the Riemann Hypothesis.
I am not quite sure what you mean by numbers that are counter-examples to the Riemann Hypothesis. The Hypothesis itself is about the complex zeros to a complex function (the claim is that the zeros with positive real part all have real part 1/2). 
[Added later] Yes, Robin proved in 1984 that the inequality holding for all $n>5040$ is equivalent to the RH. So in principle you can disprove RH by finding a single integer which fails Robin's inequality. But thirty years later, no one has succeeded by that route.
[Added later] @user128932 Sorry, I failed to answer your point about $a,b$. So suppose $a,b$ (both $>5040$) failed to satisfy Robin's inequality, so that $\sigma(a)>e^\gamma a \ln\ln a$ and $\sigma(b)>e^\gamma b \ln\ln b$, does that mean $ab$ also fails to satisfy it? 
$\sigma(n)$ is a "multiplicative function" which means (in this context) that $\sigma(mn)=\sigma(m)\sigma(n)$ provided that $m,n$ are relatively prime. So if $a,b$ were relatively prime exceptions, we would have $\sigma(ab)=\sigma(a)\sigma(b)>e^{2\gamma}(ab)(\ln\ln a)(\ln\ln b)$. 
For $ab$ to be another exception we want $\sigma(ab)>e^\gamma ab\ln\ln(ab)$.
Now $e^\gamma>1.78$, so $e^{2\gamma}>1.78e^\gamma$, which is helpful. We have $\ln(a)\ln(b)>\ln(a)+\ln(b)=\ln(ab)$ for large $a,b$, which is also helpful, eg $\ln(5051)\ln(5059)=72.73>17.06=\ln(5051\cdot5059)$. Of course this effect is reduced for $\ln\ln$, eg $(\ln\ln 5051)(\ln\ln 5059)=4.59>2.84=\ln\ln(5051\cdot5059)$. But $(\ln\ln a)(\ln\ln b)>\ln\ln(ab)$ obviously holds in general for sufficiently large $a,b$ (and maybe for all $a,b$ I have not really thought about it). 
So yes in the $a,b$ relatively prime case it looks as though $ab$ would be another exception. The case where they are not relatively prime looks more doubtful, eg $\sigma(16)=5<8=\sigma(2)\sigma(8)$.
