Part (a) can only be solved when the probability space has an atom, that is, when there exists some measurable $K\subseteq\Omega$ such that $P(K)\gt0$ and, for every measurable $A\subseteq K$, $P(A)$ is either $0$ or $P(K)$. Consider $$X_n=n\cdot\mathbf 1_K,$$ and let $\delta=P(K)$. Then, for every event $A$, if $P(A)\lt\delta$ then $P(A\cap K)=0$ hence $E(|X_n|\cdot\mathbf 1_A)=0$ for every $n$.
Part (b) is much more classical. On the probability space $\Omega=[0,1]$ endowed with its Borel sigma-algrebra and Lebesgue measure, let $$X_n=\frac{2^n}n\cdot\mathbf 1_{(1/2^{n+1},1/2^n)}.$$
Part (a) can only be solved when the probability space has an atom, that is, when there exists some measurable $K\subseteq\Omega$ such that $P(K)\gt0$ and, for every measurable $A\subseteq K$, $P(A)$ is either $0$ or $P(K)$. Consider $$X_n=n\cdot\mathbf 1_K,$$ and let $\delta=P(K)$. Then, for every event $A$, if $P(A)\lt\delta$ then $P(A\cap K)=0$ hence $E(|X_n|\cdot\mathbf 1_A)=0$ for every $n$.
Part (b) is much more classical. On the probability space $\Omega=[0,1]$ endowed with its Borel sigma-algrebra and Lebesgue measure, let $$X_n=\frac{2^n}n\cdot\mathbf 1_{(1/2^{n+1},1/2^n)}.$$