Binomial Coefficients with fractions: $\binom {m-\frac 12}m=\frac 1{2^{2m}}\binom {2m}m$.

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} {m - 1/2 \choose m} & = {\pars{m - 1/2}! \over m!\pars{-1/2}!}= {\Gamma\pars{m + 1/2} \over m!\,\Gamma\pars{1/2}} \\[5mm] & = {1 \over m!\,\root{\pi}}\,\ \overbrace{{\root{2\pi}2^{1/2 - 2m}\,\Gamma\pars{2m} \over \Gamma\pars{m}}} ^{\ds{\color{#f00}{\large\S}}\,,\ \Gamma\pars{m + 1/2}}\,,\quad \pars{~\Gamma\pars{1 \over 2} = \root{\pi}~} \\[5mm] & = {1 \over 2^{2m - 1}}\,{\pars{2m - 1}! \over m!\pars{m - 1}!} = {1 \over 2^{2m - 1}}\,{\pars{2m}!/\pars{2m} \over m!\pars{m!/m}} \\[5mm] & = {1 \over 2^{2m}}\,{\pars{2m}! \over m!\, m!} = \color{#f00}{{1 \over 2^{2m}}{2m \choose m}} \end{align} $\ds{\color{#f00}{\large\S}:\ \Gamma\!-\!Duplication\ Formula}$. See $\ds{\mathbf{6.1.18}}$ in Abramowitz & Stegun Table.

Note that there are several useful ways to express $\ds{2m \choose m}$:

$$ {2m \choose m} = 2^{2m}{m - 1/2 \choose m} = 2^{2m}\bracks{{-1/2 \choose m}\pars{-1}^{m}} = {-1/2 \choose m}\pars{-4}^{m} = {-1/2 \choose -1/2 - m}\pars{-4}^{m} $$

The other one is quite similar to this one.

In equation $(6)$ of this answer, the Euler-Maclaurin Sum Formula was used to derive the asymptotic formula $$ \binom{n+\alpha}{n}=\frac{n^\alpha}{\Gamma(1+\alpha)}\left(1+\frac{\alpha+\alpha^2}{2n}-\frac{2\alpha+3\alpha^2-2\alpha^3-3\alpha^4}{24n^2}+O\!\left(\frac1{n^3}\right)\right)\tag{1} $$ by applying the idenity $$ \binom{n+\alpha}{n}=\prod_{k=1}^n\left(1+\frac\alpha{k}\right)\tag{2} $$ Proven in equation $(11)$ of this answer, Gauss' Multiplication Formula says $$ \prod_{k=0}^{n-1}\Gamma\!\left(x+\frac kn\right) =\sqrt{n2^{n-1}\pi^{n-1}}\frac{\Gamma(nx)}{n^{nx}}\tag{3} $$ So that $$ \begin{align} \prod_{k=0}^1\binom{m+\frac k2}{m} &=\prod_{k=0}^1\frac{\Gamma\!\left(m+1+\frac k2\right)}{m!\,\Gamma\left(1+\frac k2\right)}\\ &=\frac{\Gamma(2m+2)}{m!^2\,2^{2m}\,\Gamma(2)}\\ &=\frac1{2^{2m}}\binom{2m+1}{1}\frac{(2m)!}{(m!)^2}\\ &=\frac{2m+1}{4^m}\binom{2m}{m}\tag{4} \end{align} $$ and $$ \begin{align} \prod_{k=0}^3\binom{m+\frac k4}{m} &=\prod_{k=0}^3\frac{\Gamma\!\left(m+1+\frac k4\right)}{m!\,\Gamma\left(1+\frac k4\right)}\\ &=\frac{\Gamma(4m+4)}{m!^4\,4^{4m}\,\Gamma(4)}\\ &=\frac1{4^{4m}}\binom{4m+3}{3}\frac{(4m)!}{(m!)^4}\\ &=\frac1{256^m}\binom{4m+3}{3}\binom{4m}{m}\binom{3m}{m}\binom{2m}{m}\tag{5} \end{align} $$ so that $$ \binom{m+\frac14}{m}\binom{m+\frac34}{m}=\frac{(4m+1)(4m+3)}{3\cdot64^m}\binom{4m}{m}\binom{3m}{m}\tag{6} $$ So we can compute the product $\binom{m+\frac14}{m}\binom{m+\frac34}{m}$ in terms of integer binomimals, but I don't know of a way to compute each in such terms.