 Solution

Pairwise independent events satisfy $\mathbb{P}(A_i\cap A_j)=\mathbb{P}(A_i)\mathbb{P}(A_j)$ for every distinct pair. Independent events satisfyfor every nonempty subset $I$ of the indices. For three events this includes the triple intersection, not just the three pairwise intersections.

Write $A_i=B_i\cup C$. Disjointness gives $\mathbb{P}(A_i)=p+q$, while every intersection of two or three distinct $A_i$ is exactly $C$. Thus pairwise independence is equivalent to $q=(p+q{)}^2$. Since $p+q\ge 0$,For $0\le q\le 1/16$, this is an admissible nonnegative $p$: putting $u=\sqrt{q}\in [0,1/4]$ gives $3p+q=3u-2u^2\le 5/8<1$. The four disjoint events can therefore have the stated probabilities, with the remaining probability assigned to their complement. This also covers $q=0$.

Full independence would additionally require $q=(p+q{)}^3$. Together with $q=(p+q{)}^2$, this forces $p+q$ to be zero or one. The first case gives $p=q=0$; the second gives $q=1,p=0$. Neither allows positive $p,q$. There are no such mutually independent events with $p>0$ and $q>0$.