Explanation

(a)
$\because f(x) = x^{3} + ax^{2} + bx + c$
$\therefore f^{'}(x) = 3x^{2} + 2ax + b$
$y = f(x)$ is increasing.
$\Rightarrow f^{'}(x) \geq 0,\forall x$
And for $f^{'}(x) = 0$ should not form an interval.
$\Rightarrow 4a^{2} - 4 \times 3 \times b \leq 0$
$\Rightarrow a^{2} - 3b \leq 0$
This is true for exactly 16 ordered pairs $(a,b),\ 1 \leq a,b \leq 6$ namely (1, 1), (1, 2), (1, 3), (1, 4); (1, 5), (1, 6), (2, 2),(2, 3), (2, 4), (2, 5), (2, 6); (3, 3),(3, 4),(3, 5),(3, 6) and (4, 6).
Thus, required probability $= \frac{16}{36} = \frac{4}{9}$