MST and some facts via an example
Preamble
I'll use the following example to illustrate each statement. The graph below will be $GR$.
                         

Below is the MST $M$ of $GR$. These are the bolded edges.
                         

Statement One
The first statement is false if $A$ is an internal vertex of $M$. In that case, $M-\{A\}$ will result in a forest, which is not an MST.
Suppose we choose $A = d$, then $M-\{A\}$ will result in the following graph which is not a tree.
                         

Statement Two
The second statement is true. It is pretty easy to see why $M-\{A\}$ in this case is still an MST.
Suppose we choose $A = a$ (our only choices are $\{a,g,i\}$), then $M-\{A\}$ (left) will result in the following tree. We can compare it to $GR-\{A\}$ (right) and see that it is an MST.

Statement Three
The third statement is true.
Formal Proof of Statement Three
Proof: Let $GR[T_1]$ and $GR[T_2]$ be induced subgraphs of $GR$ and $M_1'$ and $M_2'$ are their MSTs respectfully. Let $X$ be the set of edges that connect vertices of $T_1$ and $T_2$ in $GR$ (note $e \in X$).
To show $M_1$ and $M_2$ are MSTs, we must show is $M_1 = M_1'$ and $M_2 = M_2'$.
Let $w(M) = w(M_1) + w(e) + w(M_2)$ be the cost of the MST of $GR$.
Let $w(M_1')$ and $w(M_2')$ be the cost of the MST of $GR[T_1]$ and $GR[T_2]$.
Construct the graph $M' = M_1' \cup e \cup M_2'$. This graph is the MSTs of both induced graphs and a single edge that connects them. Note, we could use any edge from $X$, however for it to be a MST, we must select the minimum cost edge of $X$. So $w(M') = w(M_1') + w(e) + w(M_2')$.
Since $w(M_1')$, $w(e)$, and $w(M_2')$ are minimal, then $w(M')$ is minimal. Since $w(M')$ is minimal, then $w(M') = w(M)$. So,
\begin{align}
w(M_1') + w(e) + w(M_2') &amp;= w(M') \\ &amp;= w(M) \\ &amp;= w(M_1) + w(e) + w(M_2). 
\end{align}
So $w(M_1') = w(M_1)$ and $w(M_2') = w(M_2)$, thus $M_1' = M_1$ and $M_2' = M_2$. Hence, $M_1$ and $M_2$ are MSTs.
Intuition of Proof
I'll pick $e = \{c,d\}$ to be the edge with weight one for this example. Below is $M-\{e\}$
                         

Let $M_1$ be the tree on the left and $M_2$ be the tree on the right. Then $T_1 = \{a,b,c\}$ is the vertex set of $M_1$ and $T_2 = \{d,e,f,g,h,i\}$ is the vertex set of $M_2$. We need to construct the induced subgraphs of $GR$ on $T_1$ and $T_2$, $GR[T_1]$ and $GR[T_2]$. Below on the left is $GR[T_1]$, and below on the right is $GR[T_2]$.
                 

Now you can see the $M_1$ is the MST of $GR[T_1]$ (similarly for $M_2$).
In the proof, we abuse the fact that the MST of the induced subgraph and the original graph are minimal. We also use the fact that $M$ must contain $e$, the smallest edge that connects $M_1$ and $M_2$. We can conclude the only MST of the induced graphs is $M_1$ (for the left) and $M_2$ (for the right).
I hope this helps!