Answer
(a) It is impossible because according to Descartes' rule, either $P(x)$ or $P(-x)$ will have at least one sign change,
which means that there will be at least one positive or negative real zero for this case.
(b)$P(x)=(x^2-x+1)(x^2+2x+2)$
(c) $P(x)=(x-1)(x^2+x-1)$
(d)$P(x)=(x^2+x-1)(x^2-2x-2)$
It must be even.
Work Step by Step
(a) A polynomial of degree 3 that has no real zeros
It is impossible because according to Descartes' rule, either $P(x)$ or $P(-x)$ will have at least one sign change,
which means that there will be at least one positive or negative real zero for this case.
(b) A polynomial of degree 4 that has no real zeros
$P(x)=(x^2-x+1)(x^2+2x+2)$ each quadratic term has only complex zeros.
(c) A polynomial of degree 3 that has three real zeros, only one of which is rational
$P(x)=(x-1)(x^2+x-1)$ one rational and two irrational real zeros.
(d) A polynomial of degree 4 that has four real zeros, none of which is rational
$P(x)=(x^2+x-1)(x^2-2x-2)$ four irrational real zeros.
What must be true about the degree of a polynomial with integer coefficients if it has no real zeros?
It must be even.
