小迪同学在计算$4×(5+1)×(5^{2}+1)$时,把4写成5-1后,得到$(5-1)×(5+1)×(5^{2}+1)= (5^{2}-1)×(5^{2}+1)= 25^{2}-1$.受此启发,在求$(2+1)×(2^{2}+1)×(2^{4}+1)×(2^{8}+1)×... ×(2^{32}+1)+1$的值时,乘$(2-1)$,得到$(2-1)×(2+1)×(2^{2}+1)×(2^{4}+1)×(2^{8}+1)×... ×(2^{32}+1)+1= (2^{2}-1)×(2^{2}+1)×(2^{4}+1)×(2^{8}+1)×... ×(2^{32}+1)+1= (2^{4}-1)×(2^{4}+1)×(2^{8}+1)×... ×(2^{32}+1)+1= 2^{64}-1+1= 2^{64}$.
阅读上面的材料,解答下列问题:
(1)计算:$(1+\frac {1}{2})×(1+\frac {1}{2^{2}})×(1+\frac {1}{2^{4}})×(1+\frac {1}{2^{8}})+\frac {1}{2^{15}}$;
(2)计算:$(1-\frac {1}{2^{2}})×(1-\frac {1}{3^{2}})×(1-\frac {1}{4^{2}})×... ×(1-\frac {1}{10^{2}})$.

(1)2 (2)$\frac {11}{20}$