Let $T_{1}, T_{2}, \dots, T_{k}$ be $k$ spanning trees of a graph $G$. For any pair of vertices $u$ and $v$, if the $u$--$v$ paths in the $k$ spanning trees are pairwise openly disjoint, then the spanning trees are called completely independent spanning trees (CISTs) of $G$. In this paper, we first prove that every 3-connected 2-outerplanar triangulated disc has two completely independent spanning trees. Next, for a 3-connected 3-outerplanar triangulated disc $G$, we provide sufficient conditions for $G$ to have two completely independent spanning trees. We provide an example of a 3-connected 4-outerplanar triangulation that does not have two completely independent spanning trees.

翻译：设$T_{1}, T_{2}, \dots, T_{k}$是图$G$的$k$棵生成树。对于任意一对顶点$u$和$v$，若这$k$棵生成树中的$u$--$v$路径两两内部顶点不相交，则称这些生成树为$G$的完全独立生成树（CISTs）。本文首先证明每个3连通2-外平面三角剖分圆盘均具有两棵完全独立生成树。随后，针对3连通3-外平面三角剖分圆盘$G$，给出了$G$存在两棵完全独立生成树的充分条件。最后，我们构造了一个不具有两棵完全独立生成树的3连通4-外平面三角剖分实例。