If we take $f(x) = mx-5$, then $f(f(0)) = -15 \Rightarrow m[m(0)-5]-5 = -15 \Rightarrow -5m = -10 \Rightarrow m =2$, or $f(x)=2x-5$. We interested in finding all $k \in \mathbb{R}$ such that the solutions of $f(x)f(k-x)>0$ lie within an interval of length $2$. We find that: $(2x-5)[2(k-x)-5] > 0$; or $0 > 4x^2 - 4kx + (10k-25)$; or $x = \frac{4k \pm \sqrt{16k^2 - 4(4)(10k-25)}}{2(4)}$; or $x = \frac{4k \pm 4\sqrt{k^2 - 10k-25}}{2(4)} = \frac{k \pm (k-5)}{2} \Rightarrow x = k - \frac{5}{2}, \frac{5}{2}$ (i). Since the roots in (i) are not included in the solution set of $f(x)f(k-x)>0$, we require $\left|k - \frac{5}{2} - \frac{5}{2}\right| = |k-5| > 2$ to ensure an interval length of at least $2$. This occurs for $k \in (-\infty, 3) \cup (7, \infty)$.