Find the remainder when $(x - 1)^{100} + (x - 2)^{200}$ is divided by $x^2 - 3x + 2$ .

Write $$(x - 1)^{100} + (x - 2)^{200}=k(x)(x-2)(x-1)+ax+b$$

SInce this is valid for all $x$ it is valid also for $$x=1: \;\;\; 1=a+b$$ and $$x=2: \;\;\; 1=a2+b$$

So $a=0$ and $b=1$.

Since $(x - 1)^{99} \equiv (x - 1)\;\mod (x - 2)$, we get that

$(x - 1)^{100}\equiv (x-1)^2\;\mod (x-2)(x-1). \quad(*)$

Since $(x-2)^{199}\equiv -1\;\mod (x-1)$, we get that

$(x-2)^{200}\equiv -(x-2)\;\mod (x-1)(x-2). \quad(**)$

So, by adding $(*)$ and $(**)$, it follows that

$(x - 1)^{100} + (x - 2)^{200} \equiv (x-1)^2-(x-2)\\\mod (x-1)(x-2),$

that is

$(x - 1)^{100} + (x - 2)^{200}\equiv x^2-3x+3\mod (x^2-3x+2)$.

Hence,

$(x - 1)^{100} + (x - 2)^{200} \equiv 1\;\mod (x^2-3x+2)$.

You are right that $(x-1)^{98}\equiv1\pmod{(x-2)}$. But that implies $$(x-1)^{100}\equiv(x-1)^2=x(x-2)+1\equiv1\pmod{x-2}.$$ More naively, as $$x-1\equiv1\pmod{x-2}$$ then $$(x-1)^{100}\equiv1^{100}=1\pmod{x-2}.$$ Similarly, $$x-2\equiv-1\pmod{x-1}$$ and $$(x-2)^{200}\equiv(-1)^{200}=1\pmod{x-1}.$$ So $(x-1)^{100}+(x-2)^{200}$ is congruent to $1$ modulo both $x-1$ and $x-2$, and so also modulo $(x-1)(x-2)$.