Can Markov inequality be used to define bounds in a meaningful way? Suppose $X\sim \text{Binomial}(100,0.5)$. Recall Markov's inequality. 
$$\Bbb{P}(X\geq 5) \leq \frac{\Bbb{E}[X]}{5}=\frac{100*0.5}{5} = 10$$
Why is this inequality valuable? Since I'm working with a probability distribution, I already know $$\Bbb{P}(X\geq5)\leq 1$$ so knowing  $$\Bbb{P}(X\geq 5) \leq  10$$ doesn't tell me anything new. 
What circumstance does this inequality give me valuable information and why doesn't that apply for this case?
 A: Because it's an inequality that is independent of the distribution it's applied to, it's a worst case bound - and for the bulk of distributions people work with, such as yours, there will be bounds or exact calculations that are far tighter.  Where it is useful, though, is in proofs, where you may not want to make more than very minimal assumptions about the distribution, in this case that the associated random variable is nonnegative, so having a worst-case bound is necessary.
The main proof where Markov's inequality is used is Chebyshev's inequality, if I recall correctly.  Markov was Chebyshev's student, and Chebyshev's inequality was originally a conjecture, which Markov proved (using his inequality) in his dissertation.
A: 
What circumstance does this inequality give me valuable information and why doesn't that apply for this case?

As it is, it shows that the decay of the survival function of any non negative integrable random variable is at most $C/x$.
In your case : $X\sim \text{Binomial}(100,0.5)$, the inequality "starts to be useful" (i.e. better than knowing that a probability is always smaller than $1$) when you replace $5$ by $100$.
Most of the time, you would use it on $|X|$ where $X$ may have positive or negative values. As @jbowman stressed it, you can use it to prove Chebyshev inequality which gives an upper bound on the probability of being far away from the mean. To prove it, you simply need to apply Markov's inequlatity to $\frac{|X-\mu|^2}{\sigma}$
In turn, Chebyshev inequality leads to very readble statement as it shows that for values from a distribution with moments of order the probability  to  lie outside the interval $(\mu - \sqrt{2}\sigma, \mu + \sqrt{2}\sigma)$ does not exceed $1/2$.
Note that a similar question was asked on math.stackexchange
