Answer
a) no
b) $x=0.71m$
c) $x=1.67m$
Work Step by Step
(a) We know that the location of zero potential should serve a distance ratio of $2:5$ because the charge at the origin has been replaced from $+q$ to $+5q$ with respect to the charge $x=1m$ and the $+5q$ is at the origin. This occurs only on either side of the $-2q$ charge, between $+5q$ and $-2q$ and at $x\gt 1m$.
(b) We know that
$V_A=K(\frac{5q}{x}-\frac{2q}{1-x})=0$
$\implies \frac{5q}{x}=\frac{2}{1-x}$
This simplifies to:
$x=0.71m$
(c) We know that
$V_B=K(\frac{5q}{x}-\frac{2q}{1-x})=0$
$\implies \frac{5q}{x}=\frac{2q}{x-1}=0$
This simplifies to:
$x=1.67m$
