What is the probability that the hand will contain the A K Q J 10 of spades. A Bridge hand consists of 13 random cards taken from a deck that holds 52
cards.


*

*What is the probability that the hand will contain the A K Q J 10 of
spades.

*What is the probability of obtaining the A K Q J 10 of at least one suit.


So far, I only know that this is the probability of getting 5 spades: P=$13!\over5!⋅8!$⋅$39!\over8!⋅31!$
\over$52!\over13!⋅39!$ and I get lost from there.
 A: 
What is the probability that a $13$-card bridge hand contains the ace, king, queen, jack, and 10 of spades?

Such a hand must contain eight of the other $52 - 5 = 47$ cards in the deck.  Hence, the desired probability is 
$$\frac{\dbinom{47}{8}}{\dbinom{52}{13}}$$

What is the probability that a $13$-card bridge hand contains the ace, king, queen, jack, and 10 of at least one suit.

Since $3 \cdot 5 = 15 > 13$, the bridge hand contains an ace, king, queen, jack, and 10 of at most two suits. 
Hands with an ace, king, queen, jack, and 10 of one suit:  Choose which suit will have the specified cards.  Choose the remaining eight cards in the hand from the remaining $47$ cards in the deck.

 $$\binom{4}{1}\binom{47}{8}$$ 

Notice that the other eight cards may include a second suit from which an ace, king, queen, jack, and 10 are drawn.  We have counted those hands twice, once for each way we could designate one of those suits as the suit containing those cards.  Therefore, we must subtract those hands that an ace, king, queen, king, jack, and 10 of two suits from the total.
Hands with the ace, king, queen, jack, and 10 of two suits:  Choose which two of the four suits will have the desired cards.  Choose which three of the $52 - 2 \cdot 5 = 42$ cards remaining in the deck will be in the hand.

  $$\binom{4}{2}\binom{42}{3}$$

Therefore, the number of bride hands that contain the ace, king, queen, jack, and 10 of at least one suit is 

 $$\binom{4}{1}\binom{47}{8} - \binom{4}{2}\binom{42}{3}$$

from which you can calculate the probability. 
A: The probability for 1) should be
$$\frac{\binom{52-5}{13-5}}{\binom{52}{13}}$$
because we have to choose just $8=13-5$ cards (see JMoravitz's comment).
As regards 2), for $1\leq k\leq 5$, the probability that hand will not contain $k$ values among A K Q J 10 is
$$p_k:=\frac{\binom{52-4k}{13}}{\binom{52}{13}}.$$
Then we use inclusion-exclusion principle:
$$p:=1+\sum_{k=1}^5(-1)^k\binom{5}{k}p_k$$
