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The minimum spanning tree cut property, in its safe-edge form, says: if an edge set $A$ is contained in some minimum spanning tree, a cut of a graph respects $A$, and $e$ is a light edge crossing a cut, then $A\cup \{e\}$ is contained in some minimum spanning tree.

Choose an MST $T$ containing $A$. If $e\in T$, nothing changes. Otherwise adding $e$ creates its unique fundamental cycle. The path in $T$ between the endpoints of $e$ must cross the chosen cut; take such an edge $f$. Since the cut respects $A$, $f\notin A$. Since $e$ is light, $w(e)\le w(f)$. The exchange $T-f+e$ is a spanning tree containing $A\cup \{e\}$, with total weight no greater than that of $T$. Minimality of $T$ forces equal optimum weight, proving the claim. Equal edge weights cause no difficulty: “safe” means belonging to some compatible optimum, not to every optimum.

A cut of a graph is a partition $(S,V\setminus S)$ into two nonempty vertex sets. An edge crosses it when its endpoints lie on opposite sides. An edge set respecting a cut contains no crossing edge. A light edge crossing a cut has minimum weight among all crossing edges; there can be several tied light edges. These definitions concern the graph's weighted edges, independently of any particular spanning tree.

True. Delete $e\in T$; the resulting two connected components define a cut of a graph. If a crossing edge $f$ had $w(f)<w(e)$, the exchange $T-e+f$ would be a spanning tree of smaller total weight, contradicting the minimum spanning tree assumption. Thus $e$ is a light edge crossing a cut for this fundamental cut. It need not be uniquely light.