To prove that the events
$$ A = \text{"the first number is odd"} $$
and $$B = \text{"the second number is even"} $$
are independent, you have to show that $P(A \cap B) = P(A)P(B)$. But we have
$$ A = \{1,3,5\} \times \{1,2,3,4,5,6\},$$
$$ B = \{1,2,3,4,5,6\} \times \{2,4,6\}, $$
and
$$ A \cap B = \{1,3,5\} \times \{2,4,6\} $$
That gives $|A| = 18$, $|B| = 18$, $|A \cap B| = 9$, so, as we are in a Laplacian space,
$$ P(A) = P(B) = \frac{18}{36} = \frac 12, \quad P(A\cap B) = \frac 9{36} = \frac 14. $$
So $P(A \cap B) = P(A)P(B)$ and $A$ and $B$ are independent.
$A$ 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.
Best Answer
Yes, this is correct. The definitions are as follow (see here).
A finite set of events $\{A_i\}$ is pairwise independent if and only if every pair of events is independent — that is, if and only if for all distinct pairs of indices $m, k$, $$ \mathrm{P}(A_m \cap A_k) = \mathrm{P}(A_m)\mathrm{P}(A_k). $$ A finite set of events is mutually independent if and only if every event is independent of any intersection of the other events — that is, if and only if for every $n$-element subset ${A_i}$, $$ \mathrm{P}\left(\bigcap_{i=1}^n A_i\right)=\prod_{i=1}^n \mathrm{P}(A_i). $$ You have shown that the events $A,B$ and $C$ are pairwise independent,but not mutually independent.