Showing undecidability
I am interpreting the word "whenever" in the definition of $T$ to mean that $T$ consists of all pairs $\langle M, w \rangle$ such that either $M$ accepts $w$ and $w^R$ or $M$ rejects $w$ and $w^R$. Please, correct me if this is wrong.

I will show a reduction from $A$ to $T$. Input $\langle M, w\rangle$ will be transformed to $\langle M', w'\rangle$ as follows:


$w' = 0w1$
$M'$ on input $w'$ strips the first bit $b_1$ and last bit $b_2$ off $w'$. Let $w$ be the remaining string. If $b_1 = 0$ then $M'$ runs $M$ on $w$ and outputs accordingly. Otherwise, if $b_1 = 1$ then $M'$ simply accepts.


I claim that $\langle M, w \rangle \in A$ if and only if $\langle M', w' \rangle \in T$:


If $M$ accepts $w$ then $M'$ on $0w1$ simulates $M$ on $w$ and accepts. Also, $M'$ on $(0w1)^R = 1w^R0$ immediately accepts, since the first bit is $1$. Thus, $\langle M', w' \rangle \in T$.
If $M$ doesn't accept $w$ then $M'$ on $0w1$ simulates $M$ on $w$ and doesn't accept. However, $M'$ on $w^R = 1w^R0$ accepts. Thus, $\langle M', w' \rangle \notin T$.