[I note that there's some lack of clarity in the question; confidence intervals apply to things like parameters, as well as means or other functions of parameters; if we're talking about intervals for data that would be other kinds of interval (prediction intervals, tolerance intervals and so on). I'll proceed as if we're discussing something like means.]
If we're sticking with typical-sized polls so we have the CLT kicking in; then we're just dealing with the variances of normally distributed quantities. It depends on the dependence (specifically, the covariance) between the quantities.
$\rm{Var}(X + Y) = \rm{Var}(X) + \rm{Var}(Y) + 2 \rm{Cov}(X,Y)$
$\rm{Var}(X - Y) = \rm{Var}(X) + \rm{Var}(Y) - 2 \rm{Cov}(X,Y)$
(that doesn't rely on normality, it's general; the meaningfulness of the resulting confidence intervals depends on normality)
The width of the confidence intervals for the proportions $X$ and $Y$ and for their sum or difference are based off their respective standard errors (the square root of the variance).
If $X$ and $Y$ are independent (based on different polls for example) then the variances add because the covariances are $0$.
So square the width of the CI's for $X$ and $Y$, add them, take the square root. That's the width of the CI for the sum or difference.
If $X$ and $Y$ are two proportions from the same poll, that is wrong, since their covariance is negative. If they add to 100% or nearly so, directly add the widths of their CIs to get the width of the difference. (For the sum, the variance will be 0 - or nearly so if they don't quite add to 100% - and the width will be a multiple of the square root of that). Estimates for the covariances can actually be calculated in general, using results for the multinomial distribution.