Answer
(a) $\pm1,\pm1/2\pm1/3,\pm1/6,\pm2,\pm2/3,\pm4,\pm4/3$
(b) 0 positive, 4, 2 or 0 negative real zeros.
Work Step by Step
(a) $p=\pm1,\pm2,\pm4,q=\pm1,\pm2,\pm3,\pm6$
$\frac{p}{q}=\pm1,\pm1/2\pm1/3,\pm1/6,\pm2,\pm2/3,\pm4,\pm4/3$
(b) $P(x)$ has zero sign changes, so there are 0 positive real zeros.
$P(-x)=6x^4-3x^3+x^2-3x+4$ and there are 4 sign changes, so there may be
4, 2 or 0 negative real zeros.