To answer question 1, I think there will always be a measure for which $T$ is ergodic.
Rotation of the unit circle $T(z) = az$ is measure preserving, (with Haar measure) and ergodic only when $a$ is not a root of unity. (Walters, P. Introduction to Ergodic Theory, Theorem 1.8).
However, we can still define a measure for which $T$ is ergodic. Say $\mu(\{a^k\}) = 1/n$ where $a$ an $n$-th root of unity, $k=0,\ldots, n-1$. Then the $T$-invariant sets either have measure 0 or 1.
The argument is not very different in the absence of periodic points. For any measure preserving system $(X,\mathcal B, \nu)$ take any element $a \in X$ and define a new measure $\mu(T^ka) orb_T(a) = 3.2^{-|k|-1}$\{T^na: n \in \mathbb Z\}$. The$T$-invariant sets either contain the orbit of Let$a$and have \mu$ be a probability measure 1; or are disjoint from the orbit of on $a$ \mathcal B / orb_T(a)$with$\mu(0 + orb_T(a)) = 1$and have measure 0zero otherwise. Then$T$is ergodic on$(X, \mathcal B/ orb_T(a), \mu)$. 1 To answer question 1, I think there will always be a measure for which$T$is ergodic. Rotation of the unit circle$T(z) = az$is measure preserving, (with Haar measure) and ergodic only when$a$is not a root of unity. (Walters, P. Introduction to Ergodic Theory, Theorem 1.8). However, we can still define a measure for which$T$is ergodic. Say$\mu(\{a^k\}) = 1/n$where$a$an$n$-th root of unity,$k=0,\ldots, n-1$. Then the$T$-invariant sets either have measure 0 or 1. When there are periodic points we can use the above argument to create an ergodic measure. The argument is not very different in the absence of periodic points. For any measure preserving system take any element$a \in X$and define a new measure$\mu(T^ka) = 3.2^{-|k|-1}$. The$T$-invariant sets either contain the orbit of$a$and have measure 1; or are disjoint from the orbit of$a\$ and have measure 0.