Random Variable with IID always Gaussian? Is there a case when we assume a random variable $\epsilon$ to be IID and assume its distribution is not gaussian?
 A: Since you call it $\epsilon$, I assume you mean an error term in some kind of regression. In that case, imagine a classifier that inputs photographs and outputs the classification of dog or cat. The classifier makes some guess, based on the photo. Then there is some error term. Some photos of dogs just look like cat photos, and some cat photos look like dog photos.
Those error terms have Bernoulli distributions, so they are not Gaussian, but I am as content to consider each error independent of the others with an equal chance of making a mistake for each classification, so the error terms are iid.
A: IID random variables are not always Gaussian
The acronym IID means "independent and identically distributed".  It refers to a property of a sequence of random variables, whereby those random variables are mutually independent, with a common marginal distribution.  If the sequence $\epsilon_1, \epsilon_2, \epsilon_3, ... \sim \text{IID Dist}$ then we have:
$$\mathbb{P}(\epsilon_1 \leqslant e_1, \epsilon_2 \leqslant e_2, ... , \epsilon_n \leqslant e_n) = \prod_{i=1}^n F(e_i),$$
where $F$ is the (common) marginal distribution function for each of the random variables in the sequence.  The distribution function $F$ can be from a Gaussian distribution, or it can be from any other distribution.  There are various contexts in which we use models with IID Gaussian error terms, and in other models, we might use IID error terms that are not Gaussian.  The latter are common in financial models, where the analyst generally wants to have fat tails in the distribution, in order to avoid underestimating the probability of extreme events.
