Exists or not ....

Let the polynomial be $f(x)$ .

By the remainder theorem, when $f(x)$ is divided by $(x-a)$ , $f(a) = r(a)$ , where $r(x)$ is the remainder.

Then, by the conditions given in the condition, we have,

When $f(x)$ is divided by $(x-8)$ , $\Rightarrow f(8) = 2$ .

When $f(x)$ is divided by $(x+2)$ , $\Rightarrow f(-2) = 3$

When $f(x)$ is divided by $(x-8)(x+2)$ the divisor has degree 2. Therefore The remainder has degree 1 or less. Let $r(x) = Ax + B$ .

$f(8) = r(8) = A(8) + B$

$2 = 8A + B$

$f(-2) = r(-2) = A(-2) + B$

$3 = -2A + B$

Solving these simultaneous equations gives $A = \frac{-1}{10}, b= \frac{14}{5}$ and $r(x) = \frac{-1}{10}x + \frac{14}{5}$