a) \(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)

We can see in R.S we have \(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2+2ab+b^2=\left(a+b\right)^2=L.S\)

So \(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)

b) \(a^2+b^2=\dfrac{1}{2}\cdot\left(\left(a+b\right)^2+\left(a-b\right)^2\right)\)

Start in the R.S we have \(\dfrac{1}{2}\cdot\left(\left(a+b\right)^2+\left(a-b\right)^2\right)=\dfrac{1}{2}\cdot\left(a^2+2ab+b^2+a^2-2ab+b^2\right)=\dfrac{1}{2}\cdot\left(2a^2+2b^2\right)\)

The same with \(\dfrac{2a^2+2b^2}{2}=a^2+b^2=L.S\)

So \(a^2+b^2=\dfrac{1}{2}\cdot\left(\left(a+b\right)^2+\left(a-b\right)^2\right)\)