Proving that $ (a \cup b)^* = (b^*(a\cup\lambda)b^*)^*$?
One of the equalities 12 tell us that now
$(u\cup v)^* = u^*(vu^*)^*$.
Applying this to the last expression we obtain
$(b\cup (a\cup\lambda))^*$.

Also $(u\cup \lambda)^* = (u^*\lambda^*)^* = 
(u^* \lambda)^* = (u^*)^* = u^*$.

Do you have associativity of $\cup$ so that $(b\cup (a\cup\lambda))^* = ((b\cup a)\cup\lambda))^*$? It should be a basic property.