Let's solve each of these limits step by step.

Problem (a)

$\lim_{{x \to \frac{1}{3}}} \frac{9x^2 - 1}{3x + 1}$

1. Substitute $x = \frac{1}{3}$:

   $9\left(\frac{1}{3}\right)^2 - 1 = 9 \cdot \frac{1}{9} - 1 = 1 - 1 = 0$

   and

   $3\left(\frac{1}{3}\right) + 1 = 1 + 1 = 2.$

   Substituting directly does not yield an indeterminate form since we only get 0 in the numerator and not in the denominator. Therefore, the limit is

   $\frac{0}{2} = 0.$

So, the answer to (a) is: $0.$

Problem (b)

$\lim_{{x \to 1}} \frac{x^3 + 1}{x^2 - 1}$

1. Substitute $x = 1$:

   $x^3 + 1 = 1^3 + 1 = 2$

   and

   $x^2 - 1 = 1^2 - 1 = 0.$

   This results in an indeterminate form $\frac{2}{0}$, so we need to simplify further. Factoring the denominator, we have

   $x^2 - 1 = (x - 1)(x + 1).$

   The expression becomes:

   $\frac{x^3 + 1}{(x - 1)(x + 1)}.$

To continue from here,