(a):for every digit, the digit after it has 2 choices. Giving us $9*2^{7}=1152$ robust numbers. (b):consider a cycle starting from 5 ,in which the distinct digits of robust numbers can run either forwards or backwards $$5,9,4,8,3,7,2,6,1$$Thus, a specific number (for example'1'), can appear at any position for exactly 2 times. $$S=(1+2+3+4+5+6+7+8+9)*22222222=999999990$$

(a) Note, that every digit, has $2$, choices, besides, the first digit, which has $9,$ hence, $9\cdot 2^7=\boxed{1152}.$