To determine which of the given statements is/are true for every undirected connected graph $G$ with the property that every spanning tree has an even weight, let's analyze each option carefully.

First, consider the nature of the weights. If every spanning tree has an even weight, this means the sum of the weights of the edges in any spanning tree will always be even.

Option A

All edges in $G$ have even weight.

If all edges in $G$ have even weights, then the sum of the weights of the edges in any subset (including any spanning tree) will also be even. This would ensure that every spanning tree has an even weight.

So, Option A implies that every spanning tree has an even weight. However, it's not necessarily the only scenario where this is true, as other configurations could also result in an even weight spanning tree. Therefore, Option A is not necessarily true for every such graph $G$.

Option B

All edges in $G$ have even weight OR all edges in $G$ have odd weight.

If all edges have even weights, then clearly every spanning tree will have an even weight, as discussed in Option A. Now, consider if all edges have odd weights. In this case, each spanning tree, having an odd number of edges (since $G$ is connected and spanning tree has $n-1$ edges where $n$ is the number of vertices), the sum of the weights will be a sum of odd numbers. The sum of an odd number of odd numbers remains odd, which contradicts the given condition that the weight of spanning trees must be even.

Thus, Option B is not true for every graph $G$, because 'all edges being odd' does not lead to all spanning trees having even weight.

Option C

In each cycle $C$ in $G$, all edges in $C$ have even weight.

This option implies that within every cycle, every edge should have an even weight. This strong condition is not necessary to ensure that every spanning tree has an even weight. Hence, Option C is also not necessarily true for every such graph $G$.

Option D

In each cycle $C$ in $G$, either all edges in $C$ have even weight OR all edges in $C$ have odd weight.

If all edges in a cycle $C$ have even weights, it contributes an even sum. If all edges in a cycle have odd weights, then the sum of the weights will be even if there are even numbers of edges. However, since spanning trees are derived from the graph and ensure connectivity without cycles, maintaining an odd or even edge structure per cycle indirectly ensures the overall evenness of spanning tree weights. This condition ensures that the sum of weights in any redundant paths isn't causing any inconsistencies regarding tree weights.

Therefore, Option D is indeed a necessary and sufficient condition to ensure that every spanning tree in $G$ has an even weight.

Based on this analysis, the correct answer is:

Option D: In each cycle $C$ in $G$, either all edges in $C$ have even weight OR all edges in $C$ have odd weight.