Q.Can two events be mutually exclusive and independent simultaneously?

Definition 12.15 (independence). Two events $A,B$ are independent exactly when

When $P(A),P(B)>0$, this is exactly equivalent to $P(B/A)=P(B)$ and $P(A/B)=P(A)$ -- knowing $A$ happened tells you nothing new about $B$'s chances, and vice versa. This is the special case of the Multiplication Theorem where the conditional term collapses to the plain probability. Mutually independent events $A_1,\ldots,A_n$ satisfy the stronger requirement that EVERY sub-collection multiplies out: $P(A_{i_1}\cap\cdots\cap A_{i_k})=P(A_{i_1})\cdots P(A_{i_k})$.

Theorem 12.8. If $A,B$ are independent, then so are (i) $\bar A,B$; (ii) $A,\bar B$; (iii) $\bar A,\bar B$. (Proved via De Morgan's law: $P(\bar A\cap\bar B)=P(\overline{A\cup B})=1-P(A\cup B)=1-[P(A)+P(B)-P(A)P(B)]=[1-P(A)][1-P(B)]=P(\bar A)P(\bar B)$.)

Theorem 12.9. For events with $P(A)\ne0,\ P(B)\ne0$: (1) mutually exclusive events can NEVER be independent, and (2) independent events can NEVER be mutually exclusive. The reason is direct: independence needs $P(A\cap B)=P(A)P(B)>0$ (a genuine positive overlap), while mutual exclusivity needs $P(A\cap B)=0$ -- the two conditions contradict each other unless one event is essentially impossible.

Independence is a probability property, mutual exclusivity is a set-theoretic property. Whether two events are independent can only be checked from their PROBABILITIES ($P(A\cap B)$ vs $P(A)P(B)$); whether they are mutually exclusive is checked from the EVENTS themselves ($A\cap B=\varnothing$ or not). Typical applications: independent coin tosses/die rolls/card draws WITH replacement; a card draw WITHOUT replacement is generally NOT independent, since removing the first card changes the composition available for the second.