Divide by $4^n$ so that the identity reads (again, for $n$ even)
$$ \sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} \frac{1}{4^n} (-1)^k = \frac{1}{2^n} \binom{n}{n/2}. \tag{1}$$
Claim 1: Select a permutation $\sigma$ of $[n]$ uniformly at random. For each cycle $w$ of $\sigma$, color $w$ red with probability $1/2$; otherwise, color it blue. This creates a colored permutation $\sigma_C$. Then $$\binom{2k}{k} \binom{2n-2k}{n-k} \frac{1}{4^n}$$ is the probability that exactly $k$ of the $n$ elements of a randomly-chosen permutation $\sigma$ are colored red. (See proof of Claim 1 below.)
Claim 2: Select a permutation $\sigma$ of $[n]$ uniformly at random. Then, if $n$ is even, $$\frac{1}{2^n} \binom{n}{n/2}$$ is the probability that $\sigma$ contains only cycles of even length. (See proof of Claim 2 below.)
Combinatorial proof of $(1)$, given Claims 1 and 2: For any colored permutation $\sigma_C$, find the smallest element of $[n]$ contained in an odd-length cycle $w$ of $\sigma_C$. Let $f(\sigma_C)$ be the colored permutation for which the color of $w$ is flipped. Then $f(f(\sigma_C)) = \sigma_C$, and $\sigma_C$ and $f(\sigma_C)$ have different parities for the number of red elements but the same probability of occurring. Thus $f$ is a sign-reversing involution on the colored permutations for which $f$ is defined. The only colored permutations $\sigma_C$ for which $f$ is not defined are those that have only even-length cycles. However, any permutation with an odd number of red elements must have at least one odd-length cycle, so the only colored permutations for which $f$ is not defined have an even number of red elements. Thus the left-hand side of $(1)$ must equal the probability of choosing a colored permutation that contains only even-length cycles. The probability of selecting one of the several colored variants of a given uncolored permutation $\sigma$, though, is that of choosing an uncolored permutation uniformly at random and obtaining $\sigma$, so the left-hand side of $(1)$ must equal the probability of selecting a permutation of $[n]$ uniformly at random and obtaining one containing only cycles of even length. Therefore,
$$\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} \frac{1}{4^n} (-1)^k = \frac{1}{2^n} \binom{n}{n/2}.$$
(Clearly, $$\sum_{k=0}^n \binom{2k}{k} \binom{2n-2k}{n-k} \frac{1}{4^n} = 1,$$
which gives another combinatorial proof of the unsigned version of $(1)$ mentioned in the question.)
Proof of Claim 1: There are $\binom{n}{k}$ ways to choose which $k$ elements of a given permutation will be red and which $n-k$ elements will be blue. Given $k$ particular elements of $[n]$, the number of ways those $k$ elements can be expressed as the product of $i$ disjoint cycles is $\left[ {k \atop i} \right]$, an unsigned Stirling number of the first kind. Thus the probability of choosing a permutation $\sigma$ that has those $k$ elements as the product of $i$ disjoint cycles and the remaining $n-k$ elements as the product of $j$ disjoint cycles is $\left[ {k \atop i} \right] \left[ {n-k \atop j}\right] /n!$, and the probability that the $i$ cycles are colored red and the $j$ cycles are colored blue as well is $\left[ {k \atop i} \right] \left[ {n-k \atop j}\right]/(2^i 2^j n!).$ Summing up, the probability that exactly $k$ of the $n$ elements in a randomly chosen permutation are colored red is
\begin{align}
\frac{\binom{n}{k}}{n!} \sum_{i=1}^k \sum_{j=1}^{n-k} \frac{\left[ {k \atop i} \right] \left[ n-k \atop j \right]}{2^i 2^j} = \frac{\binom{n}{k}}{n!} \sum_{i=1}^k \frac{\left[ {k \atop i} \right]}{2^i} \sum_{j=1}^{n-k} \frac{\left[ {n-k \atop j} \right]}{2^j}.
\end{align}
The two sums are basically the same, so we'll just do the first one.
$$\sum_{i=1}^k \frac{\left[ {k \atop i} \right]}{2^i} = \left( \frac{1}{2} \right)^{\overline{k}} = \prod_{i=0}^{k-1} \left(\frac{1}{2} + i\right) = \frac{1 (3) (5) \cdots (2k-1)}{2^k} = \frac{1 (2) (3) \cdots (2k-1)(2k)}{2^k 2^k k!} = \frac{(2k)!}{4^k k!}.$$
(The first equality is the well-known property that Stirling numbers of the first kind are used to convert rising factorial powers to ordinary powers. This property can be proved combinatorially. For example, Vol. 1 of Richard Stanley's Enumerative Combinatorics, 2nd ed., pp. 34-35 contains two such combinatorial proofs.)
Thus the probability that exactly $k$ of the $n$ elements of a randomly chosen permutation are colored red is $$\frac{\binom{n}{k}}{n!} \frac{(2k)!}{4^k k! } \frac{(2n-2k)!}{4^{n-k} (n-k)!} = \binom{2k}{k} \binom{2n-2k}{n-k} \frac{1}{4^n}.$$
Proof of Claim 2: Since there can be no odd cycles, $\sigma(1) \neq 1$. Thus there are $n-1$ choices for $\sigma(1)$. We have already chosen the element that maps to $\sigma(1)$, but otherwise there are no restrictions on the value of $\sigma(\sigma(1))$, and so we have $n-1$ choices for $\sigma(\sigma(1))$ as well.
Now $n-2$ elements are unassigned. If $\sigma(\sigma(1)) \neq 1$, then we have an open cycle. We can't assign $\sigma^3(1) = 1$, as that would close the current cycle at an odd number of elements. Also, $\sigma(1)$ and $\sigma^2(1)$ are already taken. Thus there are $n-3$ choices for the value of $\sigma^3(1)$. If $\sigma(\sigma(1)) = 1$, then we have just closed an even cycle. Selecting any unassigned element in $[n]$, say $j$, we cannot have $\sigma(j) = j$, as that would create an odd cycle, and $1$ and $\sigma(1)$ are already taken. Thus we have $n-3$ choices for $\sigma(j)$ as well.
In general, if there are $i$ elements unassigned and $i$ is even, there is either one even-length open cycle or no open cycles. If there is an open cycle, we cannot close it, and so we have $i-1$ choices for the next element in the cycle. If there is not an open cycle, we select the smallest unassigned element $j$. Since we cannot have $\sigma(j) = j$, there are $i-1$ choices for $\sigma(j)$. Either way, we have $i-1$ choices. If there are $i$ elements unassigned and $i$ is odd, though, there must always be an odd-length open cycle. Since we can close it, there are $i$ choices for the next element in the cycle.
All together, then, if $n$ is even then the number of permutations of $[n]$ that contain only cycles of even length is $$(n-1)^2 (n-3)^2 \cdots (1)^2 = \left(\frac{n!}{2^{n/2} (n/2)!}\right)^2 = \frac{n!}{2^n} \binom{n}{n/2}.$$ Thus the probability of choosing a permutation uniformly at random and obtaining one that contains only cycles of even length is $$\frac{1}{2^n} \binom{n}{n/2}.$$
(I've been thinking about this problem off and on for the two months since I first posted it. What finally broke it open for me was discovering the interpretation of the unsigned version of the identity mentioned as #60 on Richard Stanley's "Bijective Proof Problems" document.)
$(a):$
It is better to use the concept of a hand of $13$, with $\binom{52}{13}$ possible hands
$Pr = \dfrac{\binom{13}5\binom{13}{3}\binom{13}{3}\binom{13}{2}}{\binom{52}{13}}$
$(b):$
Here, the sample space is $\binom{52}{13,13,13,13}$.
[If not familiar with the multinomial distribution, it is simply equivalent to $\binom{52}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13}$]
$Pr = \dfrac{\binom{13}{5,5,2,1}\binom{39}{8,8,11,12}}{\binom{52}{13,13,13,13}}$
$(c):$
Any of $4$ players can get $4$ aces and $9$ other cards, thus
$Pr = \dfrac{4\times\binom44\binom{48}9}{\binom{52}{13}}$
Best Answer
The following holonomic proof follows Peter Paule and Carsten Schneider's 2024 report Creative Telescoping for Hypergeometric Double Sums, and was generated using their software.
Denote the inner sum $$f(r,a)=\sum_{b=0}^a\binom ab^2\binom{r+b}a=\sum_{b=0}^as(r,a,b)$$ and compute two recurrences for it:
$$(a+1)^2f(r,a)+(2a+3)(2r+1)f(r,a+1)-(a+2)^2f(r,a+2)=0$$ $$2(r+1)^2f(r+1,a)-(a^2-2ar+2r^2+2r+1)f(r,a)-(a+1)^2f(r,a+1)=0$$ The first recurrence has certificate (obtained using
Prove[]) $$g(r,a,b)=\frac{(a+1)b^2(b-a+r)(2a^2-3ab-3ar+5a+b^2+2br-4b-6r+2)}{(b-a-2)^2(b-a-1)^2}s(r,a,b)$$ in the sense that the following telescoping equation holds: $$(a+1)^2s(r,a,b)+(2a+3)(2r+1)s(r,a+1,b)-(a+2)^2s(r,a+2,b)=g(r,a,b+1)-g(r,a,b)$$ Dividing both sides by $s(r,a,b)$ results in an easily verifiable rational function identity. In exactly the same manner the second recurrence is certified by $$g(r,a,b)=-\frac{b^2(a^2-ab-3ar-a+2br+b-3r-2)}{(a-b+1)^2}s(r,a,b)$$ Using these two recurrences we now compute one for $A(r)$:Call the gnarly fourth element of the output $h(r,a)$: $$ \frac{\binom ra^2}{32(r-a+2)^2(r-a+1)^2} \left(\begin{bmatrix} 80 & 8 & -206 & 206 & -104 & -10 & 10 \\ 456 & 160 & -1095 & 843 & -214 & -25 & 7 \\ 1020 & 658 & -2182 & 1172 & -160 & -12 \\ 1158 & 1114 & -2054 & 696 & -44 \\ 708 & 916 & -924 & 152 \\ 222 & 364 & -160 \\ 28 & 56 \end{bmatrix}f(r,a) + (1+a)^2\begin{bmatrix} -80 & 152 & -98 & -10 & 10 \\ -296 & 448 & -213 & -5 & 7 \\ -428 & 486 & -156 & 2 \\ -302 & 230 & -38 \\ -104 & 40 \\ -14 \end{bmatrix}f(r,a+1)\right)$$ The matrices are shorthand for polynomials whose $r^ia^j$ coefficient is $a_{ij}$, using 0-indexing. With the help of the two smaller recurrences we can verify $$(r+1)^3\binom ra^2f(r,a)-\frac14(2r+3)(3r^2+9r+7)\binom{r+1}a^2f(r+1,a)+\frac1{16}(r+2)^3\binom{r+2}a^2f(r+2,a)=h(r,a+1)-h(r,a)$$ which telescopes to $$(r+1)^3A(r)-\frac14(2r+3)(3r^2+9r+7)A(r+1)+\frac1{16}(r+2)^3A(r+2)=0$$ But $B(r)$ satisfies the same recurrence, and the first two terms of $A(r)$ and $B(r)$ agree. This proves $A=B$.