Solutions

(i) We consider two cases.

$$\begin{aligned} x - 3 \geq 0 & \implies x - 3 = 8 \\ & \implies x = 11 \end{aligned}$$

The other case:

$$\begin{aligned} x - 3 < 0 & \implies -(x - 3) = 8 \\ & \implies -x + 3 = 8 \\ & \implies x = -5 \end{aligned}$$

(ii)

$$\begin{aligned} |x - 3| < 8 & \implies -8 < x - 3 < 8 \\ & \implies -5 < x < 11 \end{aligned}$$

(iii)

$$\begin{aligned} |x + 4| < 2 & \implies -2 < x + 4 < 2 \\ & \implies -6 < x < -2 \end{aligned}$$

(iv) The first case:

$$\begin{aligned} x \leq 0 & \implies 0 \leq - x \\ & \implies 2 \leq - x + 2 \\ & \implies 1 < 2 \leq -(x - 2) \\ & \implies 1 < |x - 2| \\ & \implies 1 < |x - 2| + |x - 1| && |x - 1| \geq 0, \forall x \in \mathbb{R} \end{aligned}$$

The second case:

$$\begin{aligned} x > 0 & \implies x - 2 > - 2, \; x - 1 > -1 \\ & \implies (x - 2) + (x - 1) > - 3 \end{aligned}$$