Solution:-

From the question,

Consider $2x-7<5x+2$

By transposing we get,

$5x-2x>-7-2$

$3x<-9$

$x<-9/3$

$x<-3$

Now, consider $5x+2\le 3x+14$

So, by transposing we get,

$5x-2x-14-2$

$3x<12$

$x<12/2$

$x<6$

As per the condition given in the question, $x\in R$.

Therefore, solution set $=[-3<x\le 6]$

Set can be represented in number line as,

Solution: –

From the question,

Consider

 $-3\le \frac{1}{2}-(2\times /3)$

$-3\le (3-4x)/6$

$-18\le (3-4x)$

So, by transposing we get,

$-18-3\le -4x$
$-21\le -4x$

$x\le 21/4$

$x\le 5\frac{1}{4}$

Now, consider

$\frac{1}{2}-(2x/3)\le 2\frac{2}{3}$

$(3-4x)/6\le 8/3$

By cross multiplication we get,

$3(3-4x)\le 48$

$9-12x\le 48$

By transposing we get,

$-12x\le 48-9$

$-12x\le 39$

$12x\ge -39$

$x\ge -39$

$x\ge -39/12$

$x\ge -3\frac{1}{4}$

As per the condition given in the question, x ∈ N.

Therefore, solution set $[-3\frac{1}{4}<x\le 5\frac{1}{2}]$.

Set can be represented in number line as