Preserve the absolute deviation when relating two normal tails

$P(X<2\mu+2\sigma^2)=P(X-2\mu<2\mu+2\sigma^2-2\mu)=P(X-2\mu<2\sigma^2)=P((X-2\mu)/\sigma<2\sigma^2/\sigma)=P(Z<2\sigma)=k$.

$E(Y-X)=E(Y)-E(X)=3\mu-2\mu=\mu$. Independence gives $\operatorname{Var}(Y-X)=\operatorname{Var}(Y)+\operatorname{Var}(X)=3\sigma^2+\sigma^2=(2\sigma)^2$. Thus $Y-X\sim N(\mu,(2\sigma)^2)$.

$\begin{aligned}&P(|Y-X-\mu|>4\sigma^2)\\ & =P(Y-X-\mu>4\sigma^2)\\ &{}+P(Y-X-\mu<-4\sigma^2).\end{aligned}$

Divide the centred difference by its standard deviation 2σ:

$\begin{aligned}&P\!\left(\frac{(Y-X)-\mu}{2\sigma}>\frac{4\sigma^2}{2\sigma}\right)\\ &{}+P\!\left(\frac{(Y-X)-\mu}{2\sigma}<-\frac{4\sigma^2}{2\sigma}\right)\\ & =P(Z>2\sigma)+P(Z<-2\sigma).\end{aligned}$

By standard-normal symmetry, $P(Z>2\sigma)+P(Z<-2\sigma)=2P(Z>2\sigma)=2(1-P(Z<2\sigma))=2(1-k)$.