2. Given the sequence $\left\{x_{n}\right\}$ satisfies $x_{n+1}=x_{n}-x_{n-1}(n \geqslant 2), x_{1}=a, x_{2}=b$, let $S_{n}=$ $x_{1}+x_{2}+\cdots+x_{n}, S_{100}=$ $\qquad$ , $x_{100}=$ $\qquad$
2. $-a, 2 b-a$.
$x_{n+1}+x_{n-1}=x_{n}, x_{n}+x_{n-2}=x_{n-1}$, so $x_{n+1}=-x_{n-2}$, i.e., $x_{m}=-x_{m+3}, x_{m+6}=x_{m} .\left\{x_{n}\right\}$ is a periodic sequence with a period of $6, x_{6 k+r}=x_{r}, x_{100}=x_{4}=-x_{1}=-a . S_{100}=\left(a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+\right.$ $\left.a_{6}\right)+\left(a_{7}+a_{8}+a_{9}+a_{10}+a_{11}+a_{12}\right)+\cdots+\left(a_{91}+a_{92}+a_{93}+a_{94}+a_{95}+a_{96}\right)+\left(a_{97}+a_{98}+\right.$ $\left.a_{99}+a_{100}\right)=16\left(a_{1}+a_{2}+a_{3}+a_{4}+a_{5}+a_{6}\right)+\left(a_{1}+a_{2}+a_{3}+a_{4}\right)=a_{2}+a_{3}=2 a_{2}-a_{1}=$ $2 b-a$ (since $a_{1}+a_{4}=a_{2}+a_{5}=a_{3}+a_{6}=0$ ), hence $x_{100}=-a, S_{100}=2 b-a$.