Step 1: Define the three events clearly.
Two fair dice are thrown, one red and one non-red.
Event $A$: The red die shows either 5 or 6.
There are 2 favorable outcomes out of 6 possible values of the red die, so:
\[ P(A)=\frac{2}{6}=\frac{1}{3} \]
Event $B$: The sum of the two dice is 7.
The possible ordered pairs giving a sum of 7 are:
\[ (1,6),(2,5),(3,4),(4,3),(5,2),(6,1) \]
There are 6 favorable outcomes out of 36 total outcomes, hence:
\[ P(B)=\frac{6}{36}=\frac{1}{6} \]
Event $C$: The sum of the two dice is 8.
The favorable outcomes are:
\[ (2,6),(3,5),(4,4),(5,3),(6,2) \]
There are 5 such outcomes, so:
\[ P(C)=\frac{5}{36} \]
Step 2: Test independence of events A and B.
Two events are independent if:
\[ P(A\cap B)=P(A)\cdot P(B) \]
Event $A\cap B$ occurs when the red die is 5 or 6 and the total sum is 7.
The outcomes satisfying both conditions are:
\[ (5,2),(6,1) \]
Thus:
\[ P(A\cap B)=\frac{2}{36}=\frac{1}{18} \]
Now compute:
\[ P(A)\cdot P(B)=\frac{1}{3}\times\frac{1}{6}=\frac{1}{18} \]
Since the two values are equal, events $A$ and $B$ are independent.
Step 3: Test independence of events A and C.
Event $A\cap C$ means the red die shows 5 or 6 and the sum is 8.
The possible outcomes are:
\[ (5,3),(6,2) \]
Hence:
\[ P(A\cap C)=\frac{2}{36}=\frac{1}{18} \]
Now calculate:
\[ P(A)\cdot P(C)=\frac{1}{3}\times\frac{5}{36}=\frac{5}{108} \]
Since $\frac{1}{18}\neq\frac{5}{108}$, events $A$ and $C$ are not independent.
Step 4: Final conclusion.
Event $A$ is independent of $B$, but not independent of $C$.
Therefore, the correct choice is:
\[ \boxed{\text{(B)}} \]