11. Given the set $A=\{1,2,3,4, \cdots, 101\}, B \subseteq A$, and the smallest element in set $B$ is an even number.
(1) If the smallest element in set $B$ is 2, and the largest element is 13, find the number of sets $B$ that satisfy the condition;
(2) Find the sum of the largest elements of all sets $B$ that satisfy the condition.
(1) Set $B$ must contain the elements 2 and 13. Since there are 10 numbers between 2 and 13, each number has 2 possible choices, so the number of sets $B$ that satisfy the condition is $2^{10}$.
(2) When the smallest element is $n=2 k(k=1,2, \cdots, 50)$, the largest element could be $n, n+1, \cdots$,
101. Among these, the number of sets where the largest element is $n$ is $2^{0}$, the number of sets where the largest element is $n+1$ is $2^{0}$, the number of sets where the largest element is $n+2$ is $2^{1}$, $\ldots$, the number of sets where the largest element is 101 is $2^{100-n}$,
then $S=n+(n+1) \cdot 2^{0}+(n+2) \cdot 2^{1}+\cdots+101 \cdot 2^{100-n}$,
$$
\begin{array}{l}
2 S=2 n+(n+1) \cdot 2^{1}+(n+2) \cdot 2^{2}+\cdots+100 \cdot 2^{100-n}+101 \cdot 2^{101-n} \\
\Rightarrow-S=1+2+2^{2}+\cdots+2^{100-n}-101 \cdot 2^{101-n} \Rightarrow S=100 \cdot 2^{101-n}+1 .
\end{array}
$$

So $T=\sum_{k=1}^{50}\left(100 \cdot 2^{101-2 k}+1\right)=100 \cdot 2^{101} \cdot \sum_{k=1}^{50} 2^{-2 k}+50$
$$
=100 \cdot 2^{101} \cdot \frac{\frac{1}{4}\left(1-\frac{1}{2^{100}}\right)}{1-\frac{1}{4}}+50=\frac{50}{3}\left(2^{102}-1\right) .
$$