11. (20 points) The rules of a coin-tossing game are: toss a coin, and get 1 point for heads, and 2 points for tails.
(1) Let the score after 5 tosses be $\xi$, find the distribution of $\xi$ and the mathematical expectation $E \xi$;
(2) Find the probability of getting exactly $n\left(n \in \mathbf{N}_{+}\right)$ points.
11. (1) The probability of scoring $\xi$ from 5 throws is
$$
P(\xi=i)=\mathrm{C}_{5}^{i-5}\left(\frac{1}{2}\right)^{5}(i=5,6, \cdots, 10) \text {. }
$$

Its distribution is shown in Table 1:
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline$\xi$ & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline$P$ & $\frac{1}{32}$ & $\frac{5}{32}$ & $\frac{5}{16}$ & $\frac{5}{16}$ & $\frac{5}{32}$ & $\frac{1}{32}$ \\
\hline
\end{tabular}

Therefore, $E \xi=\sum_{i=5}^{10} i \cdot \mathrm{C}_{5}^{i-5}\left(\frac{1}{2}\right)^{5}=\frac{15}{2}$ (points).
(2) Let $p_{n}$ denote the probability of exactly scoring $n$ points.

The only scenario where $n$ points do not appear is when $n-1$ points are scored followed by a single throw resulting in a tail.

Since the probability of "not scoring $n$ points" is $1-p_{n}$, the probability of "exactly scoring $n-1$ points" is $p_{n-1}$, and the probability of "throwing a tail" is $\frac{1}{2}$, we have:
$$
\begin{aligned}
1-p_{n} & =\frac{1}{2} p_{n-1} \\
\Rightarrow p_{n} & -\frac{2}{3}=-\frac{1}{2}\left(p_{n-1}-\frac{2}{3}\right) \\
=\cdots & =\frac{1}{6}\left(-\frac{1}{2}\right)^{n-1} .
\end{aligned}
$$

Thus, $p_{n}=\frac{1}{3}\left[2+\left(-\frac{1}{2}\right)^{n}\right]$.