I saw another question on Game theory.
My answer
for part a
the nash equlibria (T, L) and (B,R).
for part-b, Player-1's action T is strictly diominated. So Player1 never choose T.
For part c, Player-2's actions L and R are similar. So, no diominant strategy.
For part d, (M,R) and (B,R).
Are my answers correct? Please check these:)
Nash Equilibria
If player 2 will play L, player 1 should play T or M. If player 2 will play R, player 1 should play M or B. And so on. Circling payoff values in this way, we get $$ \begin{array}{|c|c|c|} \hline & L & R \\ \hline T & \boxed{1}, \boxed{1} & 0, 0 \\ M & \boxed{1}, \boxed{1} & \boxed{2}, \boxed{1} \\ B & 0, 0 & \boxed{2}, \boxed{1} \\ \hline \end{array} $$
The Nash equilibria are therefore $(T,L), (M,L), (M,R), (B,R)$, which is two additional equilibria from what you listed. Any box with both numbers circled is a Nash equilibrium.
Dominant strategies
For player 1, M is a weak dominant strategy since it always does at least as good as the other options. Another way to notice this is to note that in the row for M, both the payoffs for player 1 are circled. By the same reasoning T and B are not dominant strategies.
For player 2, there is no dominant strategy, even a weak one, because if player 1 plays T, then L is better than R, and if player 1 plays B, then R is better than L.
Dominant strategy equilibrium
Since player 2 has no dominant strategy, there can be no dominant strategy equilibrium.