Do casework on when the absolute values go from positive to negative and vice versa.

Or graph them.

If $x\geq 4$ then $x-1-2(x-4)> 3+2x\Leftrightarrow x< \frac{4}{3}$. (!) If $1\leq x< 4$ then $x-1-2(4-x)> 3+2x\Leftrightarrow x> 12$ (!) If $x< 1$ then $1-x-2(4-x)> 3+2x\Leftrightarrow x< -10$. Thus, $x< -10$.

$x \geq 4 \Rightarrow -x+7 \geq 3+2x \Rightarrow x<4/3 \rightarrow \leftarrow$ $1 \leq x<4 \Rightarrow 3x-9>3x+2 \Rightarrow x>12 \rightarrow \leftarrow$ $x<1 \Rightarrow x-7>3x+2 \Rightarrow x<-10$ Therefore, $x<-10$ $\blacksquare$

$x < -10$ because of graphing it, we see that the left side of the half-plane is shaded in, but the line that separates the whole plane is dotted, plus, the original problem states that it is just $<$ without the or equal to sign. So the answer is $x < 10.$