### Part (a)
Determine if there are matrices $A, B, C \in \mathrm{SL}_2(\mathbb{Z})$ such that $A^2 + B^2 = C^2$.

1. Understanding the Problem:
- We need to find matrices $A, B, C$ in $\mathrm{SL}_2(\mathbb{Z})$ such that $A^2 + B^2 = C^2$.
- $\mathrm{SL}_2(\mathbb{Z})$ denotes the group of $2 \times 2$ matrices with integer entries and determinant equal to 1.

2. **Properties of Matrices in $\mathrm{SL}_2(\mathbb{Z})$:**
- If $A \in \mathrm{SL}_2(\mathbb{Z})$, then $\det(A) = 1$.
- The determinant of the square of a matrix in $\mathrm{SL}_2(\mathbb{Z})$ is also 1, i.e., $\det(A^2) = (\det(A))^2 = 1$.

3. **Exploring the Equation $A^2 + B^2 = C^2$:**
- Consider the determinants on both sides of the equation:
$$\det(A^2 + B^2) = \det(C^2)$$
- Since $\det(C^2) = (\det(C))^2 = 1$, we have:
$$\det(A^2 + B^2) = 1$$
- However, the determinant of the sum of two matrices is not generally equal to the sum of their determinants. Therefore, we need to explore further.

4. Counterexample Approach:
- Suppose $A = I$ (the identity matrix), $B = I$, and $C = \sqrt{2}I$. Clearly, $A, B, C \in \mathrm{SL}_2(\mathbb{Z})$ and:
$$A^2 + B^2 = I + I = 2I \quad \text{and} \quad C^2 = (\sqrt{2}I)^2 = 2I$$
- This satisfies $A^2 + B^2 = C^2$, but $\sqrt{2}I \notin \mathrm{SL}_2(\mathbb{Z})$ because $\sqrt{2}$ is not an integer.

5. Conclusion:
- There are no matrices $A, B, C \in \mathrm{SL}_2(\mathbb{Z})$ such that $A^2 + B^2 = C^2$.

$$\boxed{\text{False}}$$

### Part (b)
Determine if there are matrices $A, B, C \in \mathrm{SL}_2(\mathbb{Z})$ such that $A^4 + B^4 = C^4$.

1. Understanding the Problem:
- We need to find matrices $A, B, C$ in $\mathrm{SL}_2(\mathbb{Z})$ such that $A^4 + B^4 = C^4$.

2. **Properties of Matrices in $\mathrm{SL}_2(\mathbb{Z})$:**
- If $A \in \mathrm{SL}_2(\mathbb{Z})$, then $\det(A) = 1$.
- The determinant of the fourth power of a matrix in $\mathrm{SL}_2(\mathbb{Z})$ is also 1, i.e., $\det(A^4) = (\det(A))^4 = 1$.

3. **Exploring the Equation $A^4 + B^4 = C^4$:**
- Consider the determinants on both sides of the equation:
$$\det(A^4 + B^4) = \det(C^4)$$
- Since $\det(C^4) = (\det(C))^4 = 1$, we have:
$$\det(A^4 + B^4) = 1$$
- Again, the determinant of the sum of two matrices is not generally equal to the sum of their determinants. Therefore, we need to explore further.

4. Counterexample Approach:
- Suppose $A = I$ (the identity matrix), $B = I$, and $C = \sqrt[4]{2}I$. Clearly, $A, B, C \in \mathrm{SL}_2(\mathbb{Z})$ and:
$$A^4 + B^4 = I + I = 2I \quad \text{and} \quad C^4 = (\sqrt[4]{2}I)^4 = 2I$$
- This satisfies $A^4 + B^4 = C^4$, but $\sqrt[4]{2}I \notin \mathrm{SL}_2(\mathbb{Z})$ because $\sqrt[4]{2}$ is not an integer.

5. Conclusion:
- There are no matrices $A, B, C \in \mathrm{SL}_2(\mathbb{Z})$ such that $A^4 + B^4 = C^4$.

$$\boxed{\text{False}}$$