What is the formal way to prove this is still a minimum spanning tree?
For part (1), denote by $w$ the weight of a tree in the original tree, and by $w'$ its weight in the new tree. Since $T$ is an MST, we have $w(T) \leq w(T')$ for every tree $T'$. Also, $w'(T) = w(T)$ and $w'(T') \geq w(T')$ for all $T'$. Therefore $w'(T) = w(T) \leq w(T') \leq w'(T')$ for all trees $T'$, that is, $T$ is an MST with respect to $w'$ as well.

Part (2) is similar to part (1), though slightly more challenging. Suppose that we decreased the weight on an edge in $T$ by an amount $\delta$. Then $w'(T) = w(T) - \delta$, and $w'(T') \geq w(T') - \delta$ for every tree $T'$. Therefore $w'(T) = w(T) - \delta \leq w(T') - \delta \leq w'(T')$ for every tree $T'$, and so $T$ is an MST with respect to $w'$ as well.

I couldn't quite understand part (3), but I think you've seen enough examples of how these proofs go.