Let's solve the given problems from the uploaded image.

Problem (a):
$\frac{2}{1 - \sqrt{2}}$

To rationalize the denominator, multiply both numerator and denominator by the conjugate of $1 - \sqrt{2}$, which is $1 + \sqrt{2}$:

$\frac{2}{1 - \sqrt{2}} \cdot \frac{1 + \sqrt{2}}{1 + \sqrt{2}} = \frac{2(1 + \sqrt{2})}{(1 - \sqrt{2})(1 + \sqrt{2})}$

Simplify the denominator using the difference of squares:

$(1 - \sqrt{2})(1 + \sqrt{2}) = 1^2 - (\sqrt{2})^2 = 1 - 2 = -1$

Now simplify the fraction:

$\frac{2(1 + \sqrt{2})}{-1} = -2(1 + \sqrt{2}) = -2 - 2\sqrt{2}$

Final Answer (a): $-2 - 2\sqrt{2}$

Problem (b):
$\frac{z - 9}{3 - \sqrt{z}}$

To rationalize the denominator, multiply both numerator and denominator by the conjugate of $3 - \sqrt{z}$, which is $3 + \sqrt{z}$:

$\frac{z - 9}{3 - \sqrt{z}} \cdot \frac{3 + \sqrt{z}}{3 + \sqrt{z}} = \frac{(z - 9)(3 + \sqrt{z})}{(3 - \sqrt{z})(3 + \sqrt{z})}$

Simplify the denominator using the difference of squares:

$(3 - \sqrt{z})(3 + \sqrt{z}) = 3^2 - (\sqrt{z})^2 = 9 - z$

So, the expression becomes:

$\frac{(z - 9)(3 + \sqrt{z})}{9 - z}$

Factor out $-(z - 9)$ from the denominator:

$9 - z = -(z - 9)$

Substitute back:

$\frac{(z - 9)(3 + \sqrt{z})}{-(z - 9)} = -(3 + \sqrt{z})$

Final Answer (b): $-3 - \sqrt{z}$

Let me know if you'd like further clarification or details.

Here are five related questions to practice:

1. Rationalize $\frac{3}{2 - \sqrt{3}}$.
2. Simplify $\frac{5 + \sqrt{7}}{2 - \sqrt{7}}$ by rationalizing the denominator.
3. Rationalize $\frac{1}{\sqrt{5} - 1}$.
4. Prove the general formula for rationalizing $\frac{1}{a - \sqrt{b}}$.
5. Simplify $\frac{4z - 1}{3 + \sqrt{z}}$ by rationalizing.

Tip: When rationalizing denominators, always look for the conjugate to simplify expressions involving square roots!