$A$ and $B$ are independent because the red and blue dice are separate dice. (If one embedded magnets in them and threw them next to each other one could question such things, but it is conventional in phrasing math problems like this to construe that as meaning they will be independent.)
Now consider $\Pr(C \mid A\ \&\ B)$ and $\Pr(C)$. If these are equal then $C$ is independent of $[A\ \&\ B]$ but that does not mean that $C$ is independent of $A$, nor that $C$ is independent of $B$. What is the conditional distribution of the outcomes given by the red and blue dice given the event $[A\ \&\ B]$? Given that event, six outcomes are equally probable:
$$
\begin{array}{c|ccc}
& 3 & 4 & 5 \\
\hline 1 & 1/6 & 1/6 & 1/6 \\
2 & 1/6 & 1/6 & 1/6
\end{array}
$$
In only one of these outcomes, the sum is $7$; therefore $\Pr(C\mid A\ \&\ B) = 1/6$.
What is the marginal (i.e. unconditional) probability of $C$? It is
\begin{align}
& \Pr\Big( \Big[ 1\ \&\ 6\Big] \text{ or } \Big[ 2\ \&\ 5\Big] \text{ or } \Big[ 3\ \&\ 4\Big] \text{ or } \Big[ 4\ \&\ 3\Big] \text{ or } \Big[ 5\ \&\ 2\Big] \text{ or } \Big[ 6\ \&\ 1\Big] \text{ or } \Big) \\[10pt]
= {} & \frac 1 {36} + \frac 1 {36} + \frac 1 {36} + \frac 1 {36} + \frac 1 {36} + \frac 1 {36} = \frac 1 6.
\end{align}
Hence we have $\Pr(C) = \Pr(C\mid A\ \&\ B)$, so $C$ is independent of $[A\ \&\ B]$.
Is $C$ independent of $A$? It is if $\Pr(C\mid A) = 1/6$; otherwise it's not.
\begin{align}
& \Pr(C\mid A) = \Pr(\text{sum is 7} \mid \text{red is 3, 4, or 5} ) = \frac{\Pr(\text{sum is 7 and $A$ is 3, 4, or 5})}{\Pr(\text{red is 3, 4, or 5})} \\[10pt]
= {} & \frac{1/36 + 1/36 + 1/36}{1/2} = \frac 1 6.
\end{align}
Therefore $A$ and $C$ are independent.
Similarly
\begin{align}
& \Pr(C\mid B) = \Pr(\text{sum is 7} \mid \text{blue is 1 or 2} ) = \frac{\Pr(\text{sum is 7 and blue is 1 or 2})}{\Pr(\text{blue is 1 or 2})} \\[10pt]
= {} & \frac{1/36+1/36}{1/3} = \frac 1 6.
\end{align}
Therefore $B$ and $C$ are independent.
Thus $A$, $B$, and $C$ are pairwise independent.
Now
\begin{align}
\Pr(A\ \&\ B\ \&\ C) = \Pr(A)\Pr(B)\Pr(C\mid A\ \&\ B) = \frac 1 2 \cdot \frac 1 3 \cdot\frac 1 6 = \frac 1 {36} = \Pr(A)\Pr(B)\Pr(C)
\end{align}
and so the three are mutually independent.
Suppose we flip two (distinguishable) coins. Let $A$ be the event that the first coin shows heads, $B$ the event that the second coin shows heads, and $C$ the event that both coins show the same. These events are pairwise independent but not mutually independent.
Best Answer
We will take the following definitions
Suppose $A_1,A_2,\ldots,A_n$ are $n$ events.
Definition 1: They are pairwise independet if $$P(A_i\cap A_j)=P(A_i)P(A_j)\; \forall 1\leq i,j\leq n\;,i\not=j$$
Definition 2: They are mutually independet if $$P(A_{i_1}\cap A_{i_2}\cap\ldots\cap A_{i_m})=P(A_{i_1})P(A_{i_2})\ldots P(A_{i_m})$$ $\forall 1\leq i_1<i_2<\ldots<i_m\leq n$ , $\forall m=2,3\ldots,n$, that is, for any combination of events you choose, satisfy the product rule.
Definition 3 They are independent if $$P(A_1\cap A_2\cap \ldots \cap A_n)=P(A_1)P(A_2)\ldots P(A_n)$$
Remark :
Pairwise independent doesn't imply mutually independent but mutually implies pairwise.
Mutually independent implies independent, but not the converse because it is possible to create a three-event example in which
$${\displaystyle \mathrm {P} (A\cap B\cap C)=\mathrm {P} (A)\mathrm {P} (B)\mathrm {P} (C)}$$ and yet no two of the three events are pairwise independent (and hence the set of events are not mutually independent) George, Glyn, "Testing for the independence of three events