= Solution
For $t\geq0$, evaluate the <survival function> directly:
$$
P(X>t)=\lambda^2\int_t^\infty xe^{-\lambda x}dx
=(1+\lambda t)e^{-\lambda t}.
$$
At $t=\mu+2\sigma=(2+2\sqrt2)/\lambda$, this becomes
$$
\boxed{P(X>\mu+2\sigma)=(3+2\sqrt2)e^{-2(1+\sqrt2)}\approx0.04662.}
$$
The parameter $\lambda$ cancels because the threshold scales with the distribution. Also $\mu-2\sigma<0$ lies outside the support, so in this particular case the two-sided event $|X-\mu|>2\sigma$ has the same <probability> as the stated upper-tail event.
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