Why not just simplify your problem
(a-b)/2+a
into
(3a-b)/2
and then use the rules for the error of 3a+b. You don't need to worry about the independency of error a to error a.
This way I get
(3 Δa + Δb)/2 = 1.4
Or using the other rule [ √(Δa2 + Δb2) ]
( √(9*Δa2 + Δb2) ) / 2...