Note that before cards 4-9 are turned, there are 4 black sides up, and after cards 1-3,10-12 are turned, there are 5 black sides up. Since turning cards 4-9 and 1-3,10-12 means turning all cards, then there are 9 black sides in total. And since before turning cards 1-6 there are 9 black sides up, then all 9 black cards correspond to 9 white-black cards. The remaining cards are 3 white-white cards. An example arrangement would be cards 1,7,8 be white-white, and the rest be white-black with each black side initially faces up.

Divide the $12$ cards into $2$ sets of $6$ cards each. $Set _1$ consisting of cards $1-6$ and $Set _2$ consisting of cards $7-12$. Let $b_i$, $w_i$ be the number of cards with Black side up, white side up initially in $Set _i$. Let's label the cards having sides Black-White as $BW$, Black-Black as $BB$, White-White as $WW$. Since turning of cards from $Set_1$ reduced the total number of cards with Black side up by $5$, we have $b_1 \geq 5 --> (1)$ Case $1$: $b_1 = 5$ Then $Set _1$ must have $5BW$ and $1WW$ type of cards. Also, $Set _2$ must have $w_2 = 2$ and $b_2 = 4$ Case 2: $b_1 = 6$ Then $Set _1$ must have $6 BB$ type of cards. This implies $w_2 = 3$ and $b_2 = 3$ Now, if you analyze carefully, subsequent turning of cards $4-9$, $1-3$, followed by $10-12$ essentially results in turning of cards $7-12$ from the initial position of cards. (This is because a card which is turned twice has the same effect as not being turned at all!). So, turning cards $7-12$ (which is essentially $Set _2$) reduced the total number of cards with Black side up by $4$ (from $9$ to $5$). So this means that $Set _1$ cannot have more than 5 cards. Thus $b_1 \leq 5 --> (2)$. From $(1)$ and $(2)$, we have that $b_1 = 5$. Thus, $Set _1$ must have $5 BW$ and $1 WW$ type of cards and ince $Set _2$ must have $2$ cards with White side up and $4$ cards with Black side up and since all these cards must turn White side up, $Set_2$ must have $2 WW$ and $4 BW$ type of cards. So in total there are $9 BW$ and $3 WW$ cards.