$T(T(v))=0$ and so $T(v)\in$ Ker$(T)$. Hence (b) is true.
Consider (a) when all $T(v)=0$. Then the kernel is all of $V$ whilst the image is only {$0$}. So (a) is not true in general.
$T(T(v))=0$ and so $T(v)\in$ Ker$(T)$. Hence (b) is true.
Consider (a) when all $T(v)=0$. Then the kernel is all of $V$ whilst the image is only {$0$}. So (a) is not true in general.