Algebra

$a+ b$ = $4$ .........................(1)

$a-b$ = $2$ .........................(2)

so, $2a$ = $6$ ......................................[(1)+(2)]

or, $a$ = $3$ , so , $b$ = $1$ and $ab$ = $3 \times 1$ = $1$

(a-b)^2=(a+b)^2-4ab

Therefore,2^2=4^2-4ab

4ab=12

Therefore, ab=3

$\begin{aligned}ab&=\left( \frac{a+b}{2}\right)^2-\left(\frac{a-b}{2}\right)^2\\&=\left(\frac{4}{2}\right)^2-\left(\frac{2}{2}\right)^2\\&=2^2-1^2\\&=\boxed{3}\end{aligned}$