Question 1021583
We use the pmf of Poisson r.v.

i.  {{{p(x) = (2^x/x!)e^(-2)}}}
The answer is p(2)+p(3)+p(4) +... = 1 - p(0) - p(1) = {{{1-e^-2-2e^-2}}} = 0.594, to 3 decimal places.  (Here {{{mu = 2}}}. 

ii.  Here {{{mu = 3*2 = 6}}}
==> {{{p(1) = (6^1/1!)e^(-6) = 6e^(-6) = 0.01487}}} to five decimal places.

iii.  Here {{{mu = (3/4)*2 = 3/2}}}
==>{{{p(0) = (1.5^0/0!)e^(-1.5) = e^(-1.5) = 0.22313}}} to five decimal places.