Problem 1594

### Part (a)

1. **Understanding the Set $S$**:
We need to find the set $S$ defined as:
$$S = \{ f \in M(X) \mid f(x + y + x) = f(x) + f(y) + f(x) \ \forall x, y \in X \}$$
Given the addition table, we need to verify the condition $f(x + y + x) = f(x) + f(y) + f(x)$ for all $x, y \in X$.

2. Analyzing the Addition Table:
The addition table for $X = \{0, a, b, c\}$ is:
$$\begin{array}{c|cccc} + & 0 & a & b & c \\ \hline 0 & 0 & a & b & c \\ a & a & 0 & c & b \\ b & b & c & 0 & a \\ c & c & b & a & 0 \\ \end{array}$$
We need to check the condition $f(x + y + x) = f(x) + f(y) + f(x)$ for all $x, y \in X$.

3. Verifying the Condition:
Let's verify the condition for each pair $(x, y)$:
- For $x = 0$:
$$f(0 + y + 0) = f(y) = f(0) + f(y) + f(0) \implies f(0) = 0$$
- For $x = a$:
$$f(a + y + a) = f(0 + y) = f(y) = f(a) + f(y) + f(a) \implies f(a) = 0$$
- For $x = b$:
$$f(b + y + b) = f(0 + y) = f(y) = f(b) + f(y) + f(b) \implies f(b) = 0$$
- For $x = c$:
$$f(c + y + c) = f(0 + y) = f(y) = f(c) + f(y) + f(c) \implies f(c) = 0$$

From the above, we see that $f(x) = 0$ for all $x \in X$. Therefore, the only function in $S$ is the zero function.

4. **Counting the Elements in $S$**:
Since the only function in $S$ is the zero function, we have:
$$|S| = 1$$

### Part (b)

1. **Understanding the Set $I$**:
We need to find the set $I$ defined as:
$$I = \{ f \in M(X) \mid f(x + x) = f(x) + f(x) \ \forall x \in X \}$$
Given the addition table, we need to verify the condition $f(x + x) = f(x) + f(x)$ for all $x \in X$.

2. Analyzing the Addition Table:
From the addition table, we have:
$$\begin{array}{c|cccc} + & 0 & a & b & c \\ \hline 0 & 0 & a & b & c \\ a & a & 0 & c & b \\ b & b & c & 0 & a \\ c & c & b & a & 0 \\ \end{array}$$
We need to check the condition $f(x + x) = f(x) + f(x)$ for all $x \in X$.

3. Verifying the Condition:
Let's verify the condition for each $x$:
- For $x = 0$:
$$f(0 + 0) = f(0) = f(0) + f(0) \implies f(0) = 0$$
- For $x = a$:
$$f(a + a) = f(0) = f(a) + f(a) \implies 0 = f(a) + f(a) \implies f(a) = 0$$
- For $x = b$:
$$f(b + b) = f(0) = f(b) + f(b) \implies 0 = f(b) + f(b) \implies f(b) = 0$$
- For $x = c$:
$$f(c + c) = f(0) = f(c) + f(c) \implies 0 = f(c) + f(c) \implies f(c) = 0$$

From the above, we see that $f(x) = 0$ for all $x \in X$. Therefore, the only function in $I$ is the zero function.

4. **Counting the Elements in $I$**:
Since the only function in $I$ is the zero function, we have:
$$|I| = 1$$

The final answer is $\boxed{ |S| = 1 }$ and $|I| = 1$.