$\displaystyle a \blacktriangledown b= \frac{a+1}{\left | a\right |+ \left | b\right |}$, or, equivalently,

$\displaystyle a \blacktriangledown b=\left ( a+1 \right ) \div ( | a |+ | b | )$

$\displaystyle \frac{4}{5} \blacktriangledown \left (-\frac{4}{5} \right )=\left ( \frac{4}{5}+1 \right ) \div \left (\left| \frac{4}{5} \right |+ \left |- \frac{4}{5}\right | \right )$

$\displaystyle = \frac{9}{5} \div \left ( \frac{4}{5} +\frac{4}{5} \right )$

$\displaystyle = \frac{9}{5} \div \frac{8}{5}$

$\displaystyle = \frac{9}{5} \times \frac{5}{8}$

$\displaystyle = \frac{9}{8}$

$\displaystyle =1 \frac{1}{8}$

$\displaystyle p(x) = \frac{\left |x+2 \right |-1}{\left |x+1 \right |-2}$, or, equivalently,

$\displaystyle p(x) =\left ( \left |x+2 \right |-1 \right ) \div \left ( \left |x+1 \right |-2 \right )$

$\displaystyle p\left (-1 \frac{1}{5} \right )=\left ( \left |-1 \frac{1}{5}+2 \right |-1 \right ) \div \left ( \left |-1 \frac{1}{5}+1 \right |-2 \right )$

$\displaystyle =\left ( \left | \frac{4}{5} \right |-1 \right ) \div \left ( \left |- \frac{1}{5} \right |-2 \right )$

$\displaystyle =\left ( \frac{4}{5} -1 \right ) \div \left ( \frac{1}{5} -2 \right )$

$\displaystyle = - \frac{1}{5} \div \left ( - \frac{9}{5} \right )$

$\displaystyle = \frac{1}{5} \div \frac{9}{5}$

$\displaystyle = \frac{1}{5} \times \frac{5}{9}$

$\displaystyle = \frac{1} {9}$

$\displaystyle a \triangleright b = \left | a- \frac{1}{2}b\right | + \left | \frac{1}{2} a+b\right |$

$\displaystyle \frac{1}{3} \triangleright 4 = \left | \frac{1}{3} - \frac{1}{2} \cdot 4\right | + \left | \frac{1}{2} \cdot \frac{1}{3} +4\right |$

$\displaystyle = \left | \frac{1}{3} - 2\right | + \left | \frac{1}{6} +4\right |$

$\displaystyle = \left | -1 \frac{2}{3} \right | + \left | 4 \frac{1}{6} \right |$

$\displaystyle = 1 \frac{2}{3} + 4 \frac{1}{6}$

$\displaystyle = 5 \frac{5}{6}$

$\displaystyle g(x)= \left | \left | 1,000- \sqrt{x}\right | - x^{3} \right |$

$\displaystyle g(16)= \left | \left | 1,000- \sqrt{16}\right | - 16^{3} \right |$

$\displaystyle = \left | \left | 1,000- 4 \right | - 16^{3} \right |$

$\displaystyle = \left | \left | 996 \right | - 16^{3} \right |$

$\displaystyle = \left | 996 - 16^{3} \right |$

$\displaystyle = \left | 996 - 4,096 \right |$

$\displaystyle = \left | -3,100 \right |$

$\displaystyle =3,100$

$\displaystyle a \blacklozenge b = \left | 2a- 2b + 5 \right |$

$\displaystyle (-4) \blacklozenge (-7) = \left | 2 (-4)- 2(-7) + 5 \right |$

$\displaystyle = \left |-8- (-14) + 5 \right |$

$\displaystyle = \left |-8+14 + 5 \right |$

$\displaystyle = \left |11 \right |$

$\displaystyle = 11$

$\displaystyle 0 \le |a|< b < |c|$, which means that $\displaystyle b$ must be positive. 

If $\displaystyle a$ is nonnegative, then $\displaystyle a = |a| < b$. If $\displaystyle a$ is negative, then it follows that $\displaystyle a < 0 < b$. Either way, $\displaystyle a < b$. Therefore, $\displaystyle b$ cannot be the least. 

We now show that we cannot eliminate $\displaystyle a$ or $\displaystyle c$ as the least.

For example, if $\displaystyle a = 4, b= 6, c = 10$, then $\displaystyle a$ is the least;  we test both statements:

$\displaystyle |a|< b < |c|$

$\displaystyle |4|< 6 < |10|$

$\displaystyle 4 < 6< 10$, which is true.

$\displaystyle a+|b| = |c|$

$\displaystyle 4+|6| = |10|$

$\displaystyle 4+6 =10$, which is also true.

If $\displaystyle a = 4, b= 6, c = - 10$, then $\displaystyle c$ is the least; we test both statements:

$\displaystyle |a|< b < |c|$

$\displaystyle |4|< 6 < |-10|$

$\displaystyle 4 < 6< 10$, which is true.

$\displaystyle a+|b| = |c|$

$\displaystyle 4+|6| = |-10|$

$\displaystyle 4+6 =10$, which is also true.

Therefore, the correct response is $\displaystyle a$ or $\displaystyle c$ only.

$\displaystyle 0 \le |a|< b$, so $\displaystyle b$ must be positive. Therefore, since $\displaystyle 0 \le |b| < c$, equivalently, $\displaystyle 0 < b < c$, so $\displaystyle c$ must be positive, and

$\displaystyle |a|< b < c$

If $\displaystyle a$ is negative or zero, it is the least of the three. If $\displaystyle a$ is positive, then the statement becomes

$\displaystyle a < b < c$,

and $\displaystyle a$ is still the least of the three. Therefore, $\displaystyle c$ must be the greatest of the three.

If $\displaystyle |x- 6| > 17$, then either $\displaystyle x -6 > 17$ or $\displaystyle x- 6 < -17$. Solve separately:

$\displaystyle x -6 > 17$

$\displaystyle x -6 + 6 > 17 + 6$

$\displaystyle x> 23$

or 

$\displaystyle x- 6 < -17$

$\displaystyle x- 6 + 6< -17 + 6$

$\displaystyle x< -11$

The solution set, in interval notation, is $\displaystyle (-\infty, -11) \cup (23, \infty)$.

$\displaystyle p = 0 \bigstar 5$

Since $\displaystyle 0 < 5$, evaluate

$\displaystyle a \bigstar b = |a - b+1|$, setting  $\displaystyle a = 0, b=5$:

$\displaystyle 0 \bigstar 5 = |0 - 5+1|$

$\displaystyle 0 \bigstar 5 = |-4| = 4$

$\displaystyle q = 0 \bigstar 0$

Since $\displaystyle 0 = 0$, then select the pattern

$\displaystyle 0 \bigstar 0 = 4$

$\displaystyle r=0 \bigstar( -5)$

Since $\displaystyle 0 > -5$, evaluate

$\displaystyle a \bigstar b = |a + b-1|$, setting $\displaystyle a = 0, b=-5$:

$\displaystyle 0 \bigstar (-5) = |0 + (-5)-1|$

$\displaystyle 0 \bigstar (-5) = |-6| = 6$

$\displaystyle p = q= 4, r = 6$, so the correct choice is that $\displaystyle p = q < r$.

$\displaystyle 0 \le |m|< n < |p|$, which means that $\displaystyle n$ must be positive. 

If $\displaystyle m$ is nonnegative, then $\displaystyle m = |m| < n$. If $\displaystyle m$ is negative, then it follows that $\displaystyle m < 0 < n$. Either way, $\displaystyle m < n$. Therefore, $\displaystyle n$ cannot be the least. 

Now examine the statemtn $\displaystyle |m|+n = p$. If $\displaystyle m = 0$, then $\displaystyle n = p$ - but we are given that $\displaystyle n$ and $\displaystyle p$ are distinct. Therefore, $\displaystyle m$ is nonzero, $\displaystyle |m| > 0$, and 

$\displaystyle |m| +n > 0 + n$

and

$\displaystyle p > n$.

$\displaystyle p$ cannot be the least either.