Since, both the events are independent.
So, P(A+B) = P(A)×P(B)
a) P( Head & Getting a 2 ) = P( head ) × P( Getting a 2 )
[tex]=\dfrac{1}{2}\times \dfrac{1}{6}\\\\=\dfrac{1}{12}[/tex]
b) P( Even number & Tail ) = P( Even number ) × P( Tail )
[tex]=\dfrac{3}{6}\times \dfrac{1}{2}\\\\= \dfrac{1}{4}[/tex]
Hence, this is the required solution.