Solutions

(a)

$$\begin{aligned} |xy| = |x| |y| & \implies |-a| = |-1 \cdot a| = |-1| |a| = |a| \end{aligned}$$

(b) In the forward direction $(\Rightarrow)$
If $\bm{a \geq 0}$

$$\begin{aligned} -b \leq a \leq b \; \wedge \; a \geq 0 \implies 0 \leq a = |a| \leq b \end{aligned}$$

If $\bm{a < 0}$

$$\begin{aligned} -b \leq a \leq b \; \wedge \; a < 0 & \implies -b \leq a \\ & \implies 0 \leq a + b \\ & \implies |a| = -a \leq b \end{aligned}$$

Thus, $-b \leq a \leq b \implies |a| \leq b$.
In the backward direction $(\Leftarrow)$
If $\bm{a \geq 0}$

$$\begin{aligned} \\ [ \; a \geq 0 \; \wedge \; |a| \leq b \; ] & \implies 0 \leq a \leq b \; \wedge \; -b \leq 0 \leq a \\ & \implies -b \leq a \leq b \end{aligned}$$

If $\bm{a < 0}$

$$\begin{aligned} a < 0 \; \wedge \; |a| \leq b & \implies -a \leq b \\ & \implies -b \leq a \\ & \implies -b \leq a < 0 < b \\ & \implies - b \leq a \leq b \end{aligned}$$

In particular,

$$\begin{aligned} |a| \leq |a| & \implies -|a| \leq a \leq |a| \end{aligned}$$

(c)

$$\begin{aligned} - |a| \leq a \leq |a| \; \wedge \; - |b| \leq b \leq |b| & \implies -|a| -|b| \leq a + b \leq |a| + |b| \\ & \implies -(|a| + |b|) \leq a + b \leq |a| + |b| \\ & \implies |a + b| \leq |a| + |b| && (b) \end{aligned}$$