### Part (a)

1. **Define the function $g$:**
We need to show that for any function $f: \mathbb{Q} \times \mathbb{Q} \rightarrow \mathbb{R}$, there exists a function $g: \mathbb{Q} \rightarrow \mathbb{R}$ such that $f(x, y) \leq g(x) + g(y)$ for all $x, y \in \mathbb{Q}$.

2. **Construct $g$:**
Define $g(x)$ as follows:
$$g(x) = \sup_{y \in \mathbb{Q}} \left( f(x, y) - g(y) \right)$$
This definition ensures that $g(x)$ is the least upper bound of the set of values $f(x, y) - g(y)$ for all $y \in \mathbb{Q}$.

3. Verify the inequality:
For any $x, y \in \mathbb{Q}$, we have:
$$f(x, y) \leq g(x) + g(y)$$
By the definition of $g(x)$, for any $y \in \mathbb{Q}$:
$$f(x, y) - g(y) \leq g(x)$$
Adding $g(y)$ to both sides, we get:
$$f(x, y) \leq g(x) + g(y)$$
Thus, the function $g$ satisfies the required inequality.

### Part (b)

1. Identify the problem:
We need to find a function $f: \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ such that there is no function $g: \mathbb{Q} \rightarrow \mathbb{R}$ satisfying $f(x, y) \leq g(x) + g(y)$ for all $x, y \in \mathbb{R}$.

2. Construct the counterexample:
Consider the function $f(x, y) = x + y$. Suppose there exists a function $g: \mathbb{Q} \rightarrow \mathbb{R}$ such that:
$$f(x, y) \leq g(x) + g(y)$$
for all $x, y \in \mathbb{R}$.

3. Analyze the growth rate:
Fix $x_0 \in \mathbb{R}$. Then for all $y \in \mathbb{R}$, we have:
$$f(x_0, y) = x_0 + y \leq g(x_0) + g(y)$$
As $y \to +\infty$, the right-hand side $g(x_0) + g(y)$ must grow at least as fast as $x_0 + y$. This implies that $g(y)$ must grow at least linearly.

4. Contradiction:
Suppose $g(y)$ grows linearly, i.e., $g(y) = ay + b$ for some constants $a$ and $b$. Then:
$$x_0 + y \leq g(x_0) + ay + b$$
This implies:
$$x_0 + y \leq g(x_0) + ay + b$$
For large $y$, this inequality cannot hold unless $a \geq 1$. However, if $a = 1$, then $g(y)$ grows exactly as fast as $y$, which contradicts the requirement that $g$ must be defined on $\mathbb{Q}$ and not necessarily on $\mathbb{R}$.

5. Conclusion:
Therefore, there is no function $g: \mathbb{Q} \rightarrow \mathbb{R}$ such that $f(x, y) \leq g(x) + g(y)$ for all $x, y \in \mathbb{R}$.

$\blacksquare$