a) By taking $x = y = \frac{1}{2}$ we have that $x + y = 1$ is an integer and so is $ax + by = \frac{1}{2}(a + b)$, since $a + b$ is even by the assumption.

b) We notice that $ax + by = a(x + y) + (b - a)y$. Assume that $x + y$ and $ax + by$ are integers. Since $a(x + y)$ is an integer, since it is a product of two integers, $(b - a)y$ must be an integer as well. But in the case $b - a = 1$ this is not possible, since $y$ is noninteger by the assumption.

a) If $a = b$ then any noninteger numbers $x$ and $y$ whose sum is an integer will be suitable as in this case $ax + ay$ is an integer, as it is a product of two integers $a$ and $x + y$. In the case $a \neq b$ we can take $x = \frac{1}{a-b}$ and $y = \frac{a-b-1}{a-b}$, as in that case $x + y = 1$ and
$$ax + by = \frac{a}{a-b} + \frac{b(a-b-1)}{a-b} = \frac{(b+1)(a-b)}{a-b} = b+1.$$

b) Assume $x + y = n$ and $ax + by = m$, where $n$ and $m$ are integers. By interpreting this as a system of equations and solving for $x$ and $y$ we obtain $x = \frac{m-bn}{a-b}$ and $y = \frac{an-m}{a-b}$ (as $a$ and $b$ have different parities we have $a - b \neq 0$). If $a - b = 1$ then these solutions are integers. Thus we can not guarantee the existence of noninteger numbers with desired properties.