I actually find the two Nash equilibria, which are the pure $(a_3,a_6)$ and the mixed
$$\left(\frac 17 a_1 + \frac 17 a_2 + \frac 57 a_3, \frac 17 a_4 + \frac 17 a_5 + \frac 57 a_6\right)$$
but I don't have any idea how to get the mixed equilibrium
$$\left(\frac 12 a_1 + \frac 12 a_2, \frac 12 a_4 + \frac 12 a_5\right)$$
The book "Algorithms for Strong Nash Equilibrium with more than two agents" by Gatti and Rocco says there are three Nash equilibria. Let me know why they could find $3$ equilibria.
Here is my solution. For the player1, the best response to player 2's action $a_6$ is $a_3$ and similarly, the best response to player 1's action $a_6$ is $a_1$ for player2. the pure Nash equilibrium is $(a_3,a_6)$. In case of Mixed strategy, $x_{1} = (p_1,q_1,1-p_1-q_1)$ and $x_2=(p_2,q_2,1-p_2-q_2)$
$U_1(a_1) = p_1 * 5 + 0 + 0=5p$
$U_1(a_2) = 0 + q_1 * 5 + 0=5q$
$U_1(a_3) = 0 + 0 + (1-p_1-q_1)*1=1-p-q$
Since $p_1=q_1$, $1-p_1-q_1=5p$ => $p_1=q_1=1/7$ and
$1-p_1-q_1=5/7$.
the same thing adopts to $x_2$. Therefore, NE1= $(1/7a_1+1/7a_2+5/7a_3,1/7a_4+1/7a_5+5/7a_6)$
But I don't have any idea to get another mixed strategy $(1/2(a_1)+1/2(a_2),1/2(a_4)+1/2(a_5))$