Computing Galois Group of $\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}$ Galois Group of $\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}$ is cumbersome to computing. Is easy to find the all four possible candidates but is cumberstone to show that they are automorphisms:
For multiplication is too long showing that
$f((a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})(a'+b'\sqrt{2}+c'\sqrt{3}+d'\sqrt{6}))=f(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})f(a'+b'\sqrt{2}+c'\sqrt{3}+d'\sqrt{6})$
For some non trivial candidate since the left hand side expands in $16$ terms. What about $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}$? there are $7$ non trivial candidates and multiplication expands in $64$ terms! 
How to do it in a tricker form?
Edit(Let me be more precise in my question): 
I know that if $f\in\Gamma(\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q})$, then $f(\sqrt{2})=\pm\sqrt{2}$, $f(\sqrt{3})=\pm\sqrt{3}$ so there are $4$ candidates for $\mathbb{Q}$-Automorphisms. Candidates are given by:
$f_1(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})=a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6}$
$f_2(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})=a-b\sqrt{2}+c\sqrt{3}-d\sqrt{6}$
$f_3(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})=a+b\sqrt{2}-c\sqrt{3}-d\sqrt{6}$
$f_4(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})=a-b\sqrt{2}-c\sqrt{3}+d\sqrt{6}$
But I need to check that they are $\mathbb{Q}$-Automorphisms in $\mathbb Q(\sqrt{2},\sqrt{3})$. All candidates $f_i$ fixes $\mathbb{Q}$ and is obvious that all preserves sums, $f_1$ obviously preserves product also since is the identity, but is cumberstone to show that $f_i(xy)=f_i(x)_if(y)$ to conclude $f_i\in\Gamma(\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q})$.

$\mathbb{Q}(\sqrt 2, \sqrt 3)$ is the splitting field of $(x^2 - 2)(x^2 - 3)$ over $\mathbb{Q}$. Hence, the extension $\mathbb{Q}(\sqrt 2, \sqrt 3) / \mathbb{Q}$ is Galois. Since $\sqrt 2$ and $\sqrt 3$ are not linearly dependent, we have $[\mathbb{Q}(\sqrt 2, \sqrt 3) : \mathbb{Q}] = 4$. Thus, the Galois group $G$ has order $4$.
We know that automorphisms in $G$ permute the roots of $x^2 - 2$ and $x^2 - 3$. It's easy to verify that
$$
\sigma = \begin{cases}
\sqrt 2 \mapsto -\sqrt 2 \\
\sqrt 3 \mapsto \sqrt 3
\end{cases}
$$
and
$$
\tau = \begin{cases}
\sqrt 2 \mapsto \sqrt 2 \\
\sqrt 3 \mapsto -\sqrt 3
\end{cases}
$$
are members of $G$. Thus, $G = \{1, \sigma, \tau, \sigma \tau\}$. I'll let you find its isomorphism type.

For $\mathbb{Q}(\sqrt 2, \sqrt 3, \sqrt 5)$, follow the same method, as it's the splitting field of $(x^2 - 2)(x^2 - 3)(x^2 - 5)$. Consider the automorphisms:
$$
\begin{cases}
\sqrt 2 \mapsto -\sqrt 2 \\
\sqrt 3 \mapsto \sqrt 3 \\
\sqrt 5 \mapsto \sqrt 5
\end{cases}
$$
and
$$
\begin{cases}
\sqrt 2 \mapsto \sqrt 2 \\
\sqrt 3 \mapsto -\sqrt 3 \\
\sqrt 5 \mapsto \sqrt 5
\end{cases}
$$
and
$$
\begin{cases}
\sqrt 2 \mapsto \sqrt 2 \\
\sqrt 3 \mapsto \sqrt 3 \\
\sqrt 5 \mapsto -\sqrt 5
\end{cases}.
$$
