1. [tex]a=2,b=3[/tex] makes it so that
[tex]\displaystyle\lim_{x\to1^-}f(x)=\lim_{x\to1}(3-x)=2[/tex]
[tex]\displaystyle\lim_{x\to1^+}f(x)=\lim_{x\to1}(2x^2+3x)=5[/tex]
and because the one-sided limits do not agree, [tex]f(x)[/tex] is not continuous at [tex]x=1[/tex] under these conditions.
2. For [tex]f(x)[/tex] to be continuous, we need both limits to match up; specifically, we require
[tex]\displaystyle\lim_{x\to1}(ax^2+bx)=a+b=2[/tex]
3. Similar to (2), we need the limits from either side of [tex]x=2[/tex] to agree:
[tex]\displaystyle\lim_{x\to2^+}f(x)=\lim_{x\to2}(5x-10)=0[/tex]
[tex]\implies\displaystyle\lim_{x\to2^-}f(x)=\lim_{x\to2}(ax^2+bx)=4a+2b=0\implies 2a+b=0[/tex]
4. [tex]f(x)[/tex] will be continuous everywhere so long as [tex]a,b[/tex] are chosen such that both [tex]a+b=2[/tex] and [tex]2a+b=0[/tex].
[tex]a+b=2\implies b=2-a\implies 2a+(2-a)=a+2=0\implies a=-2\implies b=4[/tex]
5. I'll leave this part to you.
