In the first, square both sides, then solve the resulting polynomial.
looking at the first one and applying Prove It's method.
$|x^2+x-5|> 2x+1 \Rightarrow (x^2+x-5)^2> (2x+1)^2$
$x^4+2 x^3-9 x^2-10 x+25>4 x^2+4 x+1$
$x^4+2 x^3-13 x^2-14 x+24>0$
whittling this down by checking for factors of the form $(x\pm r)$ where r divides 24 we eventually reduce this to
$(x-3) (x-1) (x+2) (x+4)>0$
checking signs from $-\infty \to \infty$ we get
$(\infty, -4) \rightarrow +$
$(-4, -2) \rightarrow -$
$(-2, 1) \rightarrow +$
$(1, 3) \rightarrow -$
$(3,\infty) \rightarrow +$
and thus our solution is
$(\infty, -4) \cup (-2, 1) \cup (3, \infty)$
Solving that is much more methodical than trying to solve the absolute value equation directly.
Apply this to the second problem and see if you can work it out.