Let's remove all the psychology from the game, and assume that Blue's and Red's strategies are independent. Let's say Blue chooses 1 with probability p, and Red chooses 1 with probability q. Then the expected value of one round is
3 * (probability both chose 1: it is pq)
+ 5 (probability Blue chose 2, Red chose 1: it is (1-p)q)
+ 6 * (probability Blue chose 1, Red chose 2: it is p(1-q))
+ 4 * (probability Blue chose 2, Red chose 2: it is (1-p)(1-q))
Now we are getting:
value = 3pq + 5(1 - p)q + 6p(1 - q) + 4(1 - p)(1 - q) =
= 3pq + 5q - 5pq + 6p - 6pq + 4 - 4p - 4q + 4pq =
= 4 - 4pq + 2p + q = (2p - 0.5)(1 - 2q) + 4.5
Now, if p = 0.25, the expected value is 4.5 (for Blue); if p is different from that, the expected value may be less, depending on q. If p > 0.25, any q > 0.5 will give resulting value less than 4.5; if p < 0.25, any q < 0.5 will give resulting value less than 4.5. It means that the optimal strategy for Blue is to choose 1 with probability 25% and 2 with probability 75%.
In very much the same way, if q = 0.5, the expected value is 4.5 (for Blue); if q > 0.5 then p < 0.25 gives value bigger than 4.5 and if q < 0.5 then p > 0.25 gives value bigger than 4.5. Since Red wants to minimize the value, the optimal strategy is q = 0.5, i.e. choose 1 and 2 with the same probability.