(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 $x=\left[ x:x\in R;x\ge 5 \right]$