a) We exhibit two examples. Let $f_i = \prod_{j \neq i} (X - a_j)$, $i = 1, \dots, n$. Since the $f_i$ never vanish simultaneously, the Lagrange type interpolation polynomial $f = \sum_{i=1}^n \frac{b_i f_i^2}{f_i(a_i)^2}$ clearly satisfies the required conditions.

Another example may be obtained by adding a suitable positive constant to the square of a suitable Lagrange interpolation polynomial; for instance, let $b$ be a positive real number less than each $b_i$, and let
$$f = b + \left( \sum_{i=1}^{n} \frac{\sqrt{b_i - b}}{f_i(a_i)} f_i \right)^2.$$

b) Proceed by induction on $n$. The base case $n=1$ being clear, let $n \ge 2$.
If some $b_i = 0$, fix such an index $i$, and apply the induction hypothesis to provide a polynomial $g$ whose roots are all real, and $g(a_j) = \frac{b_j}{a_j - a_i}$ for all $j \neq i$. The polynomial $f = (X - a_i)g$ clearly satisfies the required conditions.

If no $b_i = 0$, assume, without loss of generality, that $a_1 < \dots < a_n$, and let $I$ be the set of all indices $i$ such that $b_i b_{i+1} > 0$. For each $i$ in $I$, choose $a'_i$ and $b'_i$ such that $a_i < a'_i < a_{i+1}$ and $b_i b'_i < 0$. The roots of the Lagrange polynomial solving the interpolation problem $f(a_i) = b_i$, $i = 1, \dots, n$, and $f(a'_i) = b'_i$, $i \in I$, are then all real and the conclusion follows.