已知 $\log_2\left(4 - x^2\right) - \log_{\sqrt{2}}(x - 1) = 1$。

1. State and solve the condition for the first logarithm to be defined

For $\log_2(4 - x^2)$ to be defined, the argument must be strictly positive:
$4 - x^2 > 0$
$x^2 < 4$
$-2 < x < 2$

2. State and solve the condition for the second logarithm to be defined

For $\log_{\sqrt{2}}(x - 1)$ to be defined:
$x - 1 > 0$
$x > 1$

3. Combine the conditions to obtain the final range

Taking the intersection of $-2 < x < 2$ and $x > 1$:
$1 < x < 2$

1. Change the base of the second logarithmic term to 2.

$\log_{\sqrt{2}}(x - 1) = \frac{\log_2(x - 1)}{\log_2(\sqrt{2})} = \frac{\log_2(x - 1)}{\frac{1}{2}} = 2\log_2(x - 1)$

2. Combine the logarithmic terms into a single logarithm.

$\log_2(4 - x^2) - 2\log_2(x - 1) = 1$
$\log_2(4 - x^2) - \log_2((x - 1)^2) = 1$
$\log_2\left(\frac{4 - x^2}{(x - 1)^2}\right) = 1$

3. Convert the logarithmic equation to an algebraic equation.

$\frac{4 - x^2}{(x - 1)^2} = 2^1 = 2$

4. Simplify to form a quadratic equation.

$4 - x^2 = 2(x^2 - 2x + 1)$
$4 - x^2 = 2x^2 - 4x + 2$
$3x^2 - 4x - 2 = 0$

5. Solve the quadratic equation and reject the invalid root using the condition from part (a).

$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(3)(-2)}}{2(3)}$
$x = \frac{4 \pm \sqrt{16 + 24}}{6}$
$x = \frac{4 \pm \sqrt{40}}{6} = \frac{4 \pm 2\sqrt{10}}{6} = \frac{2 \pm \sqrt{10}}{3}$
$x \approx 1.72$ or $x \approx -0.387$
From (a), $1 < x < 2$. Since $\frac{2 - \sqrt{10}}{3} < 1$, we reject it.
Therefore, there is only one solution, $x = \frac{2 + \sqrt{10}}{3}$.