Note: I'm assuming that when you wrote $SLR$ and $LALR$ that you actually meant $SLR(1)$ and $LALR(1)$.
It is certainly the case that if a grammar shows a conflict in an $LALR(1)$ automaton, that conflict will also be present in the $SLR(1)$ automaton, because the two automata have the same states and the $LALR(1)$ lookaheads for any production are a subset of the $SLR(1)$ lookaheads.
The remaining two proposals are false, as can be demonstrated by the same counter-example.
The classic example of an $LALR(1)$ grammar which is not $SLR(1)$ (taken directly from the Dragon book) is the abstraction of a grammar which attempts to discriminate between "lvalue" and "rvalue" syntaxes. (Roughly speaking, an "lvalue" is something which can be assigned to, so-named because it can appear on the left-hand side of an assignment operator; an "rvalue" is a value which cannot be assigned to, which means that it can only appear on the right-hand side.
It's usually convenient to include the production $R\to L$, which says that "lvalues" can also appear on the right-hand side of the assignment operator. But it is precisely this fact which leads to the conflict:
$$\begin{align}S &\to L = R\\
S &\to R\\
L &\to * R\\
L &\to id\\
R &\to L\\
\end{align}$$
As the Dragon book points out, that grammar leads to an $SLR(1)$ conflict in the state reached from $\text{GOTO}(I_0, L)$. That state ($I_2$ in the Dragon book exposition) contains the items $S\to L\;\cdot = R$ and $R\to L\;\cdot$. Since $=$ is in $\text{FOLLOW}(R)$, the state has a shift-reduce conflict. In the $LALR(1)$ automaton, that conflict is resolved.
Since that's a shift-reduce conflict, it doesn't address your question, which concerns reduce-reduce conflicts. But it's easy to modify the grammar slightly in order to turn the shift-reduce conflict into a reduce-reduce conflict. All that's necessary is to introduce a new intermediate non-terminal:
$$\begin{align}S &\to L' = R\\
S &\to R\\
L' &\to L\\
L &\to * R\\
L &\to id\\
R &\to L\\
\end{align}$$
That changes the itemset for state $I_2$ to include the items $\{L'\to L\;\cdot, R\to L\;\cdot\}$, which is now an $SLR(1)$ reduce-reduce conflict, for the same reason that the previous example had a shift-reduce conflict. Also for the same reason, the $LALR(1)$ algorithm resolves that conflict using more precisely-computed lookahead sets.
That disproves your statement 2; since the "if and only if" in statement 3 requires both statement 1 and statement 2 to be true, it also disproves statement 3.
