Only answers your first question.
There are $39$ cards left for the other $3$ players and among them there are $9$ spades.
Each of them gets $13$ cards.
Number the other $3$ players and for $i=1,2,3$ let $E_i$ be the event that player $i$ gets no spades.
Applying the principle of inclusion/exclusion and symmetry we find that:$$P(E_1\cup E_2\cup E_3)=3P(E_1)-3P(E_1\cap E_2)=3\left[\frac{\binom{30}{13}}{\binom{39}{13}}-\frac{\binom{30}{4}}{\binom{39}{13}}\right]=$$$$\frac{3\left[\binom{30}{13}-\binom{30}{4}\right]}{\binom{39}{13}}\approx0.044223$$
Your answer might work, mostly, but there is an easier way. You seem to not have counted the case where the first hand gets all the hearts (or maybe the second.)
The easier way: There are $\binom{39}{13}$ subsets of the deck of size $26$ and containing all the hearts. For each of those subsets, there are $\binom{26}{13}$ ways to distribute those cards to Player 1 and Player 2. And there are $\binom{26}{13}$ ways to divy up the rest of the cards to players 3 and 4.
So we've stumbled on the identity:
$$\sum_{i=0}^{13}\binom{13}{i}\binom{39}{13-i}\binom{26+i}{i}=\binom{39}{13}\binom{26}{13},$$ with the left side your count (corrected to include $i=0,$) and the right side is mine, both excluding the common factor of $\binom{26}{13}.$
If $4n$ is the number of cards in the deck, and there are $k$ cards of type $S,$ then the we get an identity:
$$\sum_{i=0}^{k}\binom{k}{i}\binom{4n-k}{n-i}\binom{3n-k+i}{i+n-k}=\binom{4n-k}{2n-k}\binom{2n}{n}.$$
As noted in comments, this identity can be proved by Vandermonde's identity.
Even more generally, if player $i$ gets $c_i$ cards, and $S$ has $k$ cards, and $c=c_1+c_2+c_3+c_4,$ then there are $\binom{c-k}{c_1+c_2-k}$ ways to pick $c_1+c_2$ cards that have all of $S,$ and $\binom{c_1+c_2}{c_1}$ ways of dividing the cards up for players $1$ and $2.$ Your count gives:
$$\sum_{i=0}^{k}\binom{k}{i}\binom{c-k}{c_1-i}\binom{c+i-k-c_1}{c_2+i-k}=\binom{c}{c_1+c_2-k}\binom{c_1+c_2}{c_1}.$$
Best Answer
There's just one little typo in your working; it should be $\frac{39!}{(13!)^3}$ and not $\frac{39!}{(13!)^4}$. Otherwise it is correct.
Another, perhaps simpler way is to just consider 13 spades and 39 non-spades. There are then $\binom{52}{13}$ ways to deal the cards, 4 of which have one player with all the spades, for a probability of $\frac{4\cdot13!\cdot39!}{52!}$.