Solution:

(a) There are four attributes, and there are three choices for each attribute. In a **Set** deck, there is one card for each combination of choices, so there are \(3^4=\boxed{81}\) cards.

(b) Note that if we choose any two cards, there is exactly one card that will complete a *set*. For example, consider the following two cards:

Since the first card has three shapes and the second card has one shape, the third card must have two shapes. Since both cards are red, the third card must also be red, and so on. Each attribute of the third card is determined uniquely by the first two cards, which gives us the following *set*:

There are \(\dbinom{81}{2}\) ways to choose two cards, and two cards uniquely generate a *set*. However, for each *set*, there are 3 ways that we could have chosen the two cards that generate the *set*. Therefore, the total number of possible *sets* is \(\dfrac{1}{3}\dbinom{81}{2}=\boxed{1080}.\)

(c) We want all three cards to be different with respect to every attribute.

There are \(81 \) ways to choose the first card in the *set*. Then for the second card, there are 2 ways to choose each attribute (since every attribute must be different from the first card), so there are \(2^4=16\) ways to choose the second card. Then the third card is uniquely determined (since every attribute must be different from the first two cards).

This gives us a count of \(81\cdot16=1296.\) But the order of the cards does not matter, so there are \(\dfrac{1296}{3!}=\boxed{216}\) *sets *in this case.

(d) There is one attribute where all three cards are the same, and they are different for every other attribute.

There are 4 ways to choose the attribute for which the cards are the same, then 3 options for this attribute. There are \(3^3=27\) ways to choose the first card (one attribute has already been determined), then \(2^3=8\) ways to choose the second card, and then the third card is uniquely determined.

So there are \(\dfrac{4\cdot3\cdot27\cdot8}{3!}=\boxed{432}\) *sets* in this case.

(e) There are two attributes where all three cards are the same, and they are different for every other attribute.

There are \(\dbinom{4}{2}=6\) ways to choose the two attributes for which they are the same, then 3 options for these attributes. There are \(3^2=9\) ways to choose the first card, then \(2^2=4\) ways to choose the second card, and then the third card is uniquely determined.

So there are \(\dfrac{6\cdot3^2\cdot9\cdot4}{3!}=\boxed{324}\) *sets *in this case.

(f) There are three attributes where all three cards are the same, and they are different for the remaining attribute.

There are \(\dbinom{4}{3}=4\) ways to choose the three attributes for which they are the same, then 3 options for these attributes. There are exactly three cards that meet these conditions, which form a *set*.

So there are \(4\cdot3^3=\boxed{108}\) *sets *in this case.

As a check, the answers for parts (c)-(f) add up to \(216+432+324+108=1080\) which is the answer for part (b).