If $U$ is the universal set and $A \cup B \cup C = U$,then ${(A - B) \cup (B - C) \cup (C - A)}'$ is equal to:

If $A=\{x \in R: \sqrt{x^2-8x+15} \in R\}$ and $B=\{x \in R: \frac{x-3}{2x-5} < \frac{x-6}{2x-11}\}$,then $A \cap B=$

If $A, B$ and $C$ are three sets such that $A \cap B = A \cap C$ and $A \cup B = A \cup C$,then:

$A$ two-digit number $\overline{ab}$ is called almost prime if one obtains a two-digit prime number by changing at most one of its digits $a$ or $b$. (For example,$18$ is an almost prime number because $13$ is a prime number). The number of almost prime two-digit numbers is:

For a real number $x$,$[x]$ denotes the greatest integer less than or equal to $x$. Then the value of $\left[\frac{1}{2}\right] + \left[\frac{1}{2} + \frac{1}{100}\right] + \left[\frac{1}{2} + \frac{2}{100}\right] + \left[\frac{1}{2} + \frac{3}{100}\right] + \ldots + \left[\frac{1}{2} + \frac{99}{100}\right] = $

Let $x_k$ be real numbers such that $x_k \geq k^4+k^2+1$ for $1 \leq k \leq 2018$. Denote $N=\sum_{k=1}^{2018} k$. Consider the following inequalities.
$I$. $\left(\sum_{k=1}^{2018} k x_k\right)^2 \leq N\left(\sum_{k=1}^{2018} k x_k^2\right)$
$II$. $\left(\sum_{k=1}^{2018} k x_k\right)^2 \leq N\left(\sum_{k=1}^{2018} k^2 x_k^2\right)$
Then,