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There are four conditions for a pair of a set and operation being a group:
1) the operation has to be associative
2) the operation has to have closure over the set
3) the set has to have an identity under the operation
4) the operation has to have inverses over the set

For all these problems we get the associative operator part for free since addition, multiplication, and function composition are associative.

1. $S = \{ f | f: \mathbb{R} \rightarrow \mathbb{R} \text{ and } f(3)=0 \}$

**Addition:**
Closure: $(f+g)(3)=f(3)+g(3)=0+0=0$, so we have closure
Identity: Is $f(x)=0$ in the set? Yes! since $f(3)=0$
Inverses: $f^{-1}(x)=-f(x)$. Is it in the set? Yes since, $f^{-1}(3)=-f(3)=-0=0$.
Therefore, S is a group under addition.

**Multiplication:**
Closure: $(f \cdot g)(3) = f(3) \cdot g(3) = 0 \cdot 0 = 0$
Identity: Is $f(x)=1$ in the set? No! $f(3)=1 \ne 0$
Therefore, S is not a group under multiplication.

**Composition:**
Closure: $(f \circ g)(3) = f(g(3)) = f(0) = \text{?}$. We aren't guarenteed anything about what $f(0)$ might be. For instance, $h(x)=1 - \frac{1}{3}x$ is in the set and $h(0) = 1$, so the result might be 1 instead of our expected 0. So, the set is not closed under composition.
Therefore, S is not a group under composition.

Here is how I worked out the first problem. The other problems are similar, but I can post solutions for them if that would help.