how to find $\Gamma(n+3/2)$
First of all the factorial, as it firstly encountered in the sense of counting the total number of arrangements possible with $n$ elements, is only defined for positive integers enlarged to $n=0$ by the convention $0!=1$. Thus, regarding to this intepretation something like $(1/2)!$ does not make any sense.
However, invoking the Gamma Function as extension of the factorial we are able to not only go beyond natural numbers but also to accept real or even complex numbers as arguments. Anyway this new, generalised factorial is in fact not the same as our original. Sure, we still get the relation
$$n!\stackrel{\color{red}{n\in\mathbb N}}{=}\Gamma(n+1)$$
But, as it is marked, this only holds for positive integer $n$. So of course, one could roughly speaking state that
$$\left(n+\frac12\right)!"="\Gamma\left(n+\frac12+1\right)=\Gamma\left(n+\frac32\right)$$
Note that this is exactly what WolframAlpha returned to your input. The latter one can be further simplified in two ways: either consider the integral representation of the Gamma Function an apply integration by parts repeatedly or by using the double factorial. Both will lead to the same solution.
I Using the integral representation of the Gamma Function
We want to find $\Gamma(n+3/2)$. Therefore we may write
\begin{align} \Gamma\left(n+\frac32\right)&=\int_0^\infty t^{n+\frac32-1}e^{-t}\mathrm dt\\ &=\int_0^\infty t^{n+\frac12}e^{-t}\mathrm dt\\ &=\left(n+\frac12\right)\int_0^\infty t^{n-\frac12}e^{-t}\mathrm dt\\ &=\left(n+\frac12\right)\left(n-\frac12\right)\int_0^\infty t^{(n-1)-\frac12}e^{-t}\mathrm dt\\ &~~\vdots\\ &=\underbrace{\left(n+\frac12\right)\left(n-\frac12\right)\cdots\left(\frac12\right)}_{=P}\underbrace{\int_0^\infty t^{-\frac12}e^{-t}\mathrm dt}_{=I}\\ \end{align}
Now we are left with the product $P$ and the integral $I$. Enforcing the substitution $t\mapsto\sqrt t$ within $I$ we get
$$I=\int_0^\infty t^{-\frac12}e^{-t}\mathrm dt=2\int_0^\infty e^{-t^2}\mathrm dt=\sqrt\pi$$
The product $P$ can be rewritten as
\begin{align} P&=\left(n+\frac12\right)\left(n-\frac12\right)\cdots\left(\frac32\right)\left(\frac12\right)\\ &=\left(\frac{2n+1}2\right)\left(\frac{2n-1}2\right)\cdots\left(\frac32\right)\left(\frac12\right)\\ &=\frac1{2^{n+1}}(2n+1)(2n-1)\cdots(3)(1)\\ &=\frac1{2^{n+1}}\frac{\color{red}{(2n+2)}(2n+1)\color{red}{(2n)}(2n-1)\cdots\color{red}{(4)}(3)\color{red}{(2)}(1)}{\color{red}{(2n+2)}\color{red}{(2n)}\cdots\color{red}{(4)}\color{red}{(2)}}\\ &=\frac1{2^{n+11}}\frac{(2n+2)!}{2^{n+1}(n+1)(n)\cdots(2)(1)}\\ &=\frac1{4^{n+1}}\frac{(2n+2)!}{(n+1)!} \end{align}
$$\therefore~\Gamma\left(n+\frac32\right)~=~\int_0^\infty t^{n+\frac12}e^{-t}\mathrm dt~=~\frac1{4^{n+1}}\frac{(2n+2)!}{(n+1)!}\sqrt\pi$$
II Using the Double Factorial
The Double Factorial is defined as
$$n!!=\begin{cases}(n)(n-2)\cdots(4)(2)&,\text{ if $n$ is even}\\(n)(n-2)\cdots(3)(1)&,\text{ if $n$ is odd}\end{cases}$$
Fortunately we can express the double factorial using the standard factorial as the following
$$(2n-1)!!=\frac{(2n)!}{2^nn!}~~~~~(2n)!!=2^nn!$$
The latter formula can be proved quite easy since
$$(2n)!!=(2n)(2n-2)\cdots(4)(2)=2^n(n)(n-1)\cdots(2)(1)=2^nn!$$
Knowing this we can compute the odd double factorial aswell hence the normal factorial is composed out of the odd and the even double factorial. Therefore we may get the odd double factorial by kicking out all even double factorials. So we get
$$(2n)!=(2n-1)!!\cdot(2n)!!\Rightarrow (2n-1)!!=\frac{(2n)!}{(2n)!!}=\frac{(2n)!}{2^nn!}$$
However, again these formulae only hold for positive integer $n$. Anyway we may use these properties to simplify your given factorial. To be precise, firstly not that the following holds due the functional relation of the Gamma Function, namely $\Gamma(z+1)=z\Gamma(z)$. So we get
\begin{align} \frac{\Gamma\left(n+\frac12\right)}{\Gamma\left(\frac12\right)}&=\left(n-\frac12\right)\frac{\Gamma\left(n-\frac12\right)}{\Gamma\left(\frac12\right)}\\ &=\left(n-\frac12\right)\left(n-\frac32\right)\frac{\Gamma\left(n-\frac32\right)}{\Gamma\left(\frac12\right)}\\ &~~\vdots\\ &=\left(n-\frac12\right)\left(n-\frac32\right)\cdots\left(\frac32\right)\left(\frac12\right)\frac{\Gamma\left(\frac12\right)}{\Gamma\left(\frac12\right)}\\ &=\left(\frac{2n-1}2\right)\left(\frac{2n-3}2\right)\left(\frac32\right)\left(\frac12\right)\\ &=\frac1{2^n}(2n-1)!! \end{align}
Now we know that
$$\left(n+\frac12\right)\Gamma\left(n+\frac12\right)=\Gamma\left(n+\frac32\right)$$
Putting this together yields to
$$\Gamma\left(n+\frac32\right)=\left(n+\frac12\right)\frac{(2n-1)!!}{2^n}\Gamma\left(\frac12\right)=\frac{(2n+1)!!}{2^{n+1}}\Gamma\left(\frac12\right)$$
For $\Gamma\left(\frac12\right)$ we either have to admit the value $\sqrt\pi$ or borrow the integral representation and again enforcing the subsitution $t\mapsto\sqrt t$ so that we get
$$\Gamma\left(\frac12\right)=\int_0^\infty t^{\frac12-1}e^{-t}\mathrm dt=\int_0^\infty t^{-\frac12}e^{-t}\mathrm dt=2\int_0^\infty e^{-t^2}\mathrm dt=\sqrt\pi$$
Combining this result with the our new form of representing the double factorial we finally get
$$\therefore~\Gamma\left(n+\frac32\right)~=~\frac{(2n+1)!!}{2^{n+1}}\Gamma\left(\frac12\right)~=~\frac{(2n+2)!}{4^{n+1}(n+1)!}\sqrt\pi$$
Wikipedia gives $$ \Gamma \left({\tfrac {1}{2}}+n\right)={\frac {(2n-1)!!}{2^{n}}}\,{\sqrt {\pi }}={\frac {(2n)!}{4^{n}n!}}{\sqrt {\pi }} $$ To get $\Gamma \left(n+{\tfrac {3}{2}}\right)$, use $n+1$ instead of $n$ in this formula.