Example 2 
(1) (2004 National College Entrance Examination Hunan Paper) Two identical fair coins are tossed simultaneously. The random variable $\xi=1$ indicates that there is at least one head up, and $\xi=0$ indicates that there are no heads up. Then $E \xi=$ $\qquad$
(2) (2003 Shandong Provincial Competition Question) In a basketball game, with only 0.1 seconds left before the end, Team A gets a chance for two free throws, and at that moment, Team A is trailing 100:101. Statistical data on the free-throw shooter of the team shows: the first shot has a hit rate of 0.6. If the first shot is successful, the hit rate for the second shot is 0.8; if the first shot misses, the hit rate for the second shot is 0.7. The expected value of the points scored by the team after the free throws is $\qquad$
(1) Fill in 0.75. Reason: Let $A_{i}$ represent the event “the $i$-th coin lands heads up”, where $i=1,2$. Then there are the following four distinct random events: $\bar{A}_{1} \cdot \bar{A}_{2}, \bar{A}_{1} \cdot A_{2}, A_{1} \cdot \bar{A}_{2}, A_{1} \cdot A_{2}$.

According to the given conditions, $P\left(A_{i}\right)=\frac{1}{2}, P\left(\bar{A}_{i}\right)=\frac{1}{2}, i=1,2$, and $A_{1}$ and $A_{2}$ are independent. Thus, $P\left(\bar{A}_{1} \cdot \bar{A}_{2}\right)=\frac{1}{4}, P\left(\bar{A}_{1} \cdot A_{2}\right)=\frac{1}{4}, P\left(A_{1} \cdot \bar{A}_{2}\right)=\frac{1}{4}, P\left(A_{1} \cdot A_{2}\right)=\frac{1}{4}$. Therefore, $P(\xi=1)=P\left(A_{1} \cdot \bar{A}_{2}\right)+P\left(\bar{A}_{1} \cdot A_{2}\right)+P\left(A_{1} \cdot A_{2}\right)=\frac{3}{4}, P(\xi=0)=P\left(\bar{A}_{1} \cdot \bar{A}_{2}\right)=\frac{1}{4}$.
Hence, $E \xi=1 \times 0.75+0 \times 0.25=0.75$.
(2) Fill in 101.36. Reason: Let $\xi$ denote the final score of the team, then $P(\xi=100)=0.4 \times 0.3=0.12, P(\xi=101)=0.6 \times 0.2+0.4 \times 0.7=0.4, P(\xi=102)=0.6 \times 0.8=0.48$. Therefore, $E(\xi)=100 \times 0.12+101 \times 0.4+102 \times 0.48=101.36$.