Not sure what you mean by "cough". I assume you mean the number on the face.
The rolls are independent, so
[tex]\mathbb P(\text{first roll < 4 AND second roll > 4})=\mathbb P(\text{first roll < 4})\cdot\mathbb P(\text{second roll > 4})[/tex]
For the first roll, there are three ways of getting a number less than 4 (1, 2, or 3) with a probability of [tex]\dfrac36=\dfrac12[/tex].
For the second roll, there are two ways of getting a number greater than 4 (5 or 6) with a probability of [tex]\dfrac26=\dfrac13[/tex].
So the probability of both events occurring in the prescribed order is [tex]\dfrac12\cdot\dfrac13=\dfrac16[/tex].
