The Question: 
The Attempt: I am not sure what they mean by "random variables". Am I finding an example, whether it is continuous or not, which satisfies these two inequalities?
This question has to do with finding estimators for any types of random variables.
Thank you very much!
The hint in the comments certainly provides a reasonable example here, but I wanted to stress that there are also examples using random variables with finite moments. I'll give two examples.
Example 1 - Infinite Variance
We will let $X$ be a shifted t-distribution centered at 1. $$X \sim t(\nu=2,\mu=1)$$ $$Y \sim N(\mu=1, \sigma=10)$$
For a t-distribution with $\nu$ degrees of freedom, only the first $\nu-1$ moments are defined. To simplify things, let $X' = X-1$ and $Y'=Y-1$ so that $X'\sim t(2)$ and $Y'\sim N(0,10)$. (For details on these calculations see the Folded Distributions). $$E|X-1| = E|X'| = 4f_X(0|\nu=2) \approx 1.414214$$ $$E|Y-1| = E|Y'| = 10\sqrt{\frac{2}{\pi}} \approx 7.978846$$
On the other hand, simply notice that the moments of $Y$ are well defined (finite) yet the second moment of the t-distribution with $\nu = 2$ is $\infty$.
Example 2 - Finite Moments
There are plenty of examples which do not require $X$ to have infinite variance. Consider a variable $X$ which is equal to 1 with probability $p_x$ and equal to some constant $c_x > 1$ with probability $1-p_x$. Do the same with $Y$ choosing $p_y$ and $c_y > 1$.
We simply need to choose $p_x$, $p_y$, $c_x$ and $c_y$ such that: $$(c_y-1)(1-p_y) > (c_x-1)(1-p_x)$$ $$(c_y-1)^2(1-p_y) < (c_x-1)^2(1-p_x)$$
For example, choosing $$P(X=x) = \begin{cases} 0.99, & x=1 \\ 0.01, & x = 101 \end{cases}$$ $$P(Y=y) = \begin{cases} 0.5, & x=1 \\ 0.5, & x = 11 \end{cases}$$ does the trick. The intuition is that $X$ will be close to $1$ with high probability, but when it misses it misses by a lot. Indeed we get:
$$E|X-1| = 1 < E|Y-1| = 5$$ $$E((X-1)^2) = 100 > E((Y-1)^2) = 50$$