For $(a)$, the number of such hands can be calculated as follows:$$X=\text{number of all possible hands}\\-\text{number of all possible hands with no picture card}\\=\binom{52}{13}-\binom{52-4\times 4}{13}$$
For $(b)$, we choose a 4-tuple of same-rank cards in $13$ different ways, leaving the other cards to be chosen independently; totally$$Y=13\times\binom{52-4}{13-4}$$
Edit
In case you want exactly 1 4-tuple, the answer is $$Y=13\times\binom{48}{9}-\binom{13}{2}\times\binom{44}{5}$$
For $(c)$, it is simply $$Z=4^{13}$$(why?)
I’m assuming that you mean that the five cards can be divided into three face cards, a king and a spade (and not just that they contain three cards, contain a king and contain a spade, possibly overlapping).
Let’s count separately depending on whether the spade is also a face card.
If the spade is a face card, we have $5$ face cards, of which at least one is a spade and at least one is a king. There are $\binom{12}5$ hands with $5$ face cards, of which $\binom95$ have no spade, $\binom85$ have no king and $\binom65$ have neither, so by inclusion–exclusion that makes $\binom{12}5-\binom95-\binom85+\binom65$. But now we’ve counted hands with only the king of spades and no other king and no other spade, of which there are $\binom64$, so we have to subtract those.
If the spade is not a face card, we have one of $10$ non-face-card spades, and $4$ face cards, of which one is a king. There are $\binom{12}4$ sets of $4$ face cards, of which $\binom84$ have no king, so that makes $10\left(\binom{12}4-\binom84\right)$.
Thus, in total, there are $\binom{12}5-\binom95-\binom85+\binom65-\binom64+10\left(\binom{12}4-\binom84\right)=4851$ such hands.
Since this doesn’t agree with another answer that’s been posted, I wrote Java code to check it by enumeration.
Best Answer
A five-card hand contains at least a pair unless it contains five cards of different ranks. There are $13$ ranks in the deck. The number of ways of selecting five different ranks is $\binom{13}{5}$. For each rank, we must select one of the four suits, which can be done in $\binom{4}{1}$ ways. Hence, there are $$\binom{13}{5}\binom{4}{1}^5$$ ways to select five cards from different ranks.
Subtracting this quantity from the total number of ways of selecting five cards from the $52$ cards in the deck yields the number of five-card hands which do not contain a pair. $$\binom{52}{5} - \binom{13}{5}\binom{4}{1}^5$$