If [tex]1-4x\ge0[/tex], then by definition of absolute value
[tex]|1-4x| = 1-4x > 7 \implies 4x < -6 \implies x < -\dfrac32[/tex]
On the other hand, if [tex]1-4x<0[/tex], then
[tex]|1-4x| = -(1-4x) = -1 + 4x > 7 \implies 4x > 8 \implies x > 2[/tex]
So, the overall solution to the inequality is interval union
[tex]\left(x < -\dfrac32\right) \cup (x > 2)[/tex]