Answer
(a) $(-\infty,3]$
(b) $(-\infty,-2)U(3,\infty) $
Work Step by Step
(a) $3x-7\le 8-2x \longrightarrow 5x\le 15 \longrightarrow x\le 3 $ or $(-\infty,3]$
(b) $x^2-x-6\gt0 \longrightarrow (x+2)(x-3)\gt0 \longrightarrow $ boundary points $x=-2,3$ and possible intervals $(-\infty,-2),(-2,3),(3,\infty) \longrightarrow $ choose test values $x=-3,0,4$ and get results $True,\ False,\ True \longrightarrow $ solutions $(-\infty,-2)U(3,\infty) $
