The outcome of selecting the card is independent of (not effected by) the outcome of rolling the dice.

If two events, $A$ and $B$, are independent then $P(A\cap B)=P(A)\times P(B)$.

Part a)

Let $A$ represent the event that a $\var{suit}$ is selected and let $B$ represent the event that a number greater than $\var{n}$ is rolled.

 $P(A)$ is $\frac{13}{52}=\frac{1}{4}$ and $P(B)$ is  $\frac{\var{6-n}}{6}$.

Therefore the probability of drawing a $\var{suit}$ and rolling a number greater than $\var{n}$ is $P(A) \times P(B)=\frac{1}{4} \times \frac{\var{6-n}}{6}$. 

Part b)

The probability that neither of these events occur is the probability of not drawing a $\var{suit}$ which is $P(A^c)=\frac{3}{4}$ mulitiplied by the probability of not rolling a number greater than $\var{n}$ which is $P(B^c)=\frac{\var{n}}{6}$ .

Part c)

The probability that only one of these events occur is $P(A ^c\cap B)+P(A \cap B^c)=(\frac{3}{4} \times \frac{\var{6-n}}{6})+(\frac{1}{4} \times \frac{\var{n}}{6})$.

Part d)

The probability  that at least one of these two events will occur is $1- P$(neither of the events occur)$=1-(P(A^c)\times P(B^c))= 1-(\frac{3}{4}\times \frac{\var{n}}{6})$