(i)

$3x\ge 12$

By cross multiplication we get,

$x\ge 12/3$

$x\ge 4$

As per the condition given in the question, the replacement set is R.

Therefore, solution set $x=\left[ x:x\in R;x\ge 4 \right]$

(ii)

$2x-3\ge 7$

By transposing we get,

$2x-3\ge 7$

$2x>7+3$

$2x>10$

$x>10/2$

$x>5$

As per the condition given in the question, the replacement set is R.

Therefore, solution set $x=\left[ x:x\in R;x\ge 5 \right]$