= Solution
With a cyclically oriented <independence digraph of events>, each event is independent of one of the other two. Independence is symmetric, so all three pairs are <independent events>. This does not imply mutual independence. Put $x=\mathbb P(A_1\cap A_2\cap A_3)$. The <inclusion-exclusion principle> gives
$$
\mathbb P(A_1^c\cap A_2^c\cap A_3^c)=1-3a+3a^2-x\geq1-3a+2a^2=(1-a)(1-2a),
$$
because $x\leq\mathbb P(A_1\cap A_2)=a^2$. Thus \b[the oriented case guarantees positive avoidance exactly when $0<a<1/2$].
For sharpness, for any $1/2\leq a<1$ prescribe the law of the three event indicators by assigning mass $(1-a)^2$ to each of the three singleton success patterns, $(1-a)(2a-1)$ to each doubleton, $1-3a+3a^2$ to the tripleton, and zero to the zero-success pattern. All masses are nonnegative and their sum is one. Each marginal is
$$
(1-a)^2+2(1-a)(2a-1)+(1-3a+3a^2)=a,
$$
and each pair intersection has mass $(1-a)(2a-1)+(1-3a+3a^2)=a^2$. These events satisfy the oriented independence conditions and have zero avoidance <probability>.
The unoriented triangle imposes no independence conditions at all: every vertex is adjacent to every other. The <union bound> gives positive avoidance if $3a<1$. To see that this is sharp for every $a\geq1/3$, use uniform measure on the circle of circumference one and take three arcs of length $a$ starting at $0,1/3,2/3$. They cover the circle. Hence \b[the unoriented case guarantees positive avoidance exactly when $0<a<1/3$].
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