Consider first two inequalities,

$$\begin{aligned} & \frac{4}{x+1} \leq 3 \\ & \Rightarrow \quad 4 \leq 3(x+1) \\ & \Rightarrow \quad 4 \leq 3 x+3 \\ & \Rightarrow \quad 4-3 \leq 3 x \quad \text{[subtracting 3 on both sides]}\\ & \Rightarrow \quad 1 \leq 3 x \\ & \therefore \quad x \geq \frac{1}{3}\quad \supset \text{(i)} \end{aligned}$$

and consider last two inequalities,

$$\begin{array}{rlr} & 3 \leq \frac{6}{x+1} \\ \Rightarrow & 3(x+1) \leq 6 \\ \Rightarrow & 3 x+3 \leq 6 \\ \Rightarrow & 3 x \leq 6-3 \quad \text{[subtracting 3 to both sides]}\\ \Rightarrow & 3 x \leq 3 \quad \text{[dividing by 3]}\\ \therefore & x \leq 1 \quad \supset \text{(ii)} \end{array}$$

From Eqs. (i) and (ii),

$$\begin{aligned} & x \in\left[\frac{1}{3}, 1\right] \\ & \frac{1}{3} \leq x \leq 1 \end{aligned}$$

$\begin{aligned}\text{Let}\quad |x-2| & =y \\ \frac{y-1}{y-2} & \leq 0\end{aligned}$

$$\begin{aligned} & \Rightarrow \quad y-1=0 \text { and } y-2=0 \\ & \Rightarrow \quad y=1 \text { and } y=2 \end{aligned}$$

$$\begin{array}{lc} \Rightarrow & 1 \leq y<2 \\ \Rightarrow & 1 \leq|x-2|<2 \\ \Rightarrow & 1 \leq|x-2| \text { and }|x-2|<2 \\ \Rightarrow & x-2 \leq-1 \\ \Rightarrow & x-2 \geq 1 \\ \text { and } & -2< x-2<2 \\ \Rightarrow & x \leq 1 \text { or } x \geq 3 \text { and } 0< x < 4 \\ \Rightarrow & x \in(0,1] \cup[3,4) \end{array}$$

Given, $$\frac{1}{|x|-3} \leq \frac{1}{2}$$

$$\begin{array}{lrr} \Rightarrow & |x|-3 \geq 2 & {\left[\because \frac{1}{a}<\frac{1}{b} \Rightarrow a>b\right]} \\ \Rightarrow & |x| \geq 5 & \text { [adding } 3 \text { to both sides] } \\ \Rightarrow & x \leq-5 \text { or } x \geq 5 & {[\because|x| \geq a \Rightarrow|x| \leq-a \Rightarrow|x| \geq a]} \end{array}$$

$$\begin{array}{ll} \Rightarrow & x \in(-\infty,-5] \cup[5, \infty) \quad \supset \text{(i)}\\ \text { But } & |x|-3 \neq 0 \\ \text { Either } & |x|-3<0 \text { or }|x|-3>0 \\ \Rightarrow & |x|<3 \text { or }|x|>3 \\ \Rightarrow & -3< x<3 \text { or } x<-3 \text { or } x>3 \quad \supset \text{(ii)} \end{array}$$

$$\begin{aligned} &[\because|x| < a \quad \Rightarrow-a< x < a \text { and }|x| > a \Rightarrow x < -a x>a]\\ &\text { On combining results of Eqs. (i) and (ii), we get }\\ &x \in(-\infty,-5] \cup(-3,3) \cup[5, \infty) \end{aligned}$$

$$\begin{array}{lr} & |x-1| \leq 5 \\ & -5 \leq x-1 \leq 5 \\ \Rightarrow & -4 \leq x \leq 6 \\ \Rightarrow & x \in[-4,6]\quad \supset\text{(i)} \end{array}$$

$$ \begin{array}{lr} \text { and } & |x| \geq 2 \\ \Rightarrow & x \leq-2 \text { or } x \geq 2 \\ \Rightarrow & x \in(-\infty,-2] \cup[2, \infty)\quad \supset\text{(ii)} \end{array}$$

On combining Eqs. (i) and (ii), we get

$x\in(-4,-2]\cup [2,6]$

$$\begin{aligned} &\begin{aligned} \text { We have, } \quad & -5 \leq \frac{2-3 x}{4} \\ \Rightarrow \quad& -20 \leq 2-3 x \quad \text { [multiplying by } 4 \text { on both sides] }\\ \Rightarrow \quad& 3 x \leq 2+20 \\ \Rightarrow \quad& 3 x \leq 22 \\ \Rightarrow \quad& x \leq \frac{22}{3} \\ \text { and }\quad & \frac{2-3 x}{4} \leq 9 \end{aligned}\\ \end{aligned}$$

$$\begin{array}{lc} \Rightarrow & 2-3 x \leq 36 \\ \Rightarrow & -3 x \leq 36-2 \\ \Rightarrow & -3 x \leq 34 \\ \Rightarrow & 3 x \geq-34 \\ \Rightarrow & x \geq-\frac{34}{3} \\ \Rightarrow & -\frac{34}{3} \leq x \leq \frac{22}{3} \\ \Rightarrow & x \in\left[\frac{-34}{3}, \frac{22}{3}\right] \end{array}$$