\((\mathbf i+\mathbf j)\times(\mathbf i-\mathbf j)=(\mathbf i+\mathbf j)\times\mathbf i+(\mathbf i+\mathbf j)\times(-\mathbf j)\)
since the cross product is distributive over addition. By anticommutativity,
\((\mathbf i+\mathbf j)\times\mathbf i+(\mathbf i+\mathbf j)\times(-\mathbf j)=-(\mathbf i\times(\mathbf i+\mathbf j))-(-\mathbf j)\times(\mathbf i+\mathbf j))\)
Scalars can be factored, so
\(-(\mathbf i\times(\mathbf i+\mathbf j))-(-\mathbf j)\times(\mathbf i+\mathbf j))=-(\mathbf i\times(\mathbf i+\mathbf j))+\mathbf j\times(\mathbf i+\mathbf j))\)
Using the distributive property again,
\(-(\mathbf i\times(\mathbf i+\mathbf j))+\mathbf j\times(\mathbf i+\mathbf j))=-(\mathbf i\times\mathbf i+\mathbf i\times\mathbf j)+\mathbf j\times\mathbf i+\mathbf j\times\mathbf j\)
Any vector crossed with itself is the zero vector, so you get
\(-(\mathbf i\times\mathbf i+\mathbf i\times\mathbf j)+\mathbf j\times\mathbf i+\mathbf j\times\mathbf j=-(\mathbf i\times\mathbf j)+\mathbf j\times\mathbf i\)
Because \(\mathbf i\times\mathbf j=\mathbf k\), and by anticommutativity, you are left with
\(-(\mathbf i\times\mathbf j)+\mathbf j\times\mathbf i=-\mathbf k-(\mathbf i\times\mathbf j)=-\mathbf k-\mathbf k=-2\mathbf k\)
Next, without outlining which properties are being used,
\(\mathbf k\times(\mathbf i-2\mathbf j)=\mathbf k\times\mathbf i+2(\mathbf j\times\mathbf k)=2\mathbf i+\mathbf j\)
