= Solution
Read the lower-left entry of each cell as player 1's <payoff> and the upper-right entry as player 2's <payoff>, as indicated by the source diagram. Against $L$, player 1 strictly prefers $B$ because $2\pi>1$; against $R$, player 1 strictly prefers $T$ because $3>2$. Against $T$, player 2 strictly prefers $R$ because $6>0$; against $B$, player 2 strictly prefers $L$ because $2\sqrt2>2$. Thus the <pure strategy> <Nash equilibria> are
$$
\boxed{(T,R)\quad\text{and}\quad(B,L).}
$$
For a fully <mixed strategy> <Nash equilibrium>, write $p=\Pr(T)$ and $q=\Pr(L)$. Player 1's <expected payoffs> from the two <pure strategies> are
$$
U_1(T,q)=3-2q,\qquad
U_1(B,q)=2+(2\pi-2)q.
$$
Equality, required when both <pure strategies> have positive <probability>, gives
$$
q^*=\frac1{2\pi}.
$$
Player 2's corresponding <expected payoffs> are
$$
U_2(p,L)=2\sqrt2(1-p),\qquad
U_2(p,R)=2+4p,
$$
so equality gives
$$
p^*=\frac{\sqrt2-1}{\sqrt2+2}=\frac{3\sqrt2-4}{2}.
$$
Both probabilities lie strictly between zero and one. Hence the unique fully <mixed strategy> <Nash equilibrium> is
$$
\boxed{
\sigma_1(T)=\frac{\sqrt2-1}{\sqrt2+2},\qquad
\sigma_2(L)=\frac1{2\pi},
}
$$
with complementary probabilities on $B$ and $R$. Its <expected payoffs> are
$$
U_1=3-\frac1\pi,\qquad U_2=6\sqrt2-6.
$$
These are \b[all three <Nash equilibria>]. There are no equilibria with just one player genuinely mixing: against each opposing <pure strategy>, the other player's <best response> is unique, so that player cannot mix. When both players mix, the two linear indifference equations have the unique solution just obtained. This is <support enumeration for a bimatrix game> with all possible supports checked.
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