You can try all the possible values for x to see whether the congruence holds
(a)(i)
Put x=0,1,2,3,4,5
$\displaystyle 2x^2+3x+1\equiv 1,0,3,4,3,0\pmod{6}$
$\displaystyle x\equiv 1,5\pmod{6}$
(ii)
Put x=0,1,2,3,...,10
$\displaystyle x^3-3x^2+2\equiv 2,0,9,2,7,8,0,0,3,4,9\pmod{11}$
$\displaystyle x\equiv 1,6,7\pmod{11}$
factorize the polynomial may help but you have to solve it carefully
$\displaystyle 2x^2+3x+1\equiv (2x+1)(x+1)\equiv 0\pmod{6}$
There are 4 conditions:
$\displaystyle 2x+1\equiv 0\pmod{6}$
$\displaystyle x+1\equiv 0\pmod{6}$
$\displaystyle 2x+1\equiv 2\pmod{6},x+1\equiv 3\pmod{6}$
$\displaystyle 2x+1\equiv 3\pmod{6},x+1\equiv 2\pmod{6}$