What is the probability that two cards drawn from a deck are both face cards and at least one is red?

Your mistake is in thinking that there is a $\tfrac34$ chance to get at least $1$ red; this would be the probability when drawing with replacement. But without replacement, the color of the one card affects the possibilities for the color of the other card.

If you count the number of ways to draw two face cards, you will find there are $\tbinom{12}{2}=66$ ways. The number of ways to draw no red cards is $\binom{6}{2}=15$, so the number of ways to draw at least one red card is $66-15=51$. This means the probability of at least one of the two drawn cards being red is $\tfrac{51}{66}\neq\tfrac{3}{4}$. Your second computation also reflects this.

$3/4$ is the probability that two cards drawn with replacement are not both black. However, because the cards are drawn without replacement, you cannot use this computation (the color of the first card affects the color probabilities for the second card).