First, note that statement $4$ must always be true, since the identity subgroup is always normal. (I suspect they accidentally omitted the word 'nontrivial'.)
Second, note that it is certainly possible that $G$ contains both a unique Sylow $7$-subgroup and a unique Sylow $2$-subgroup, evidenced by the cyclic group of order $56$. So statement $3$ is false if you interpret the 'or' as exclusive (which, grammatically, is implied by the 'either'). Otherwise, we proceed as follows.
Suppose that no Sylow $7$-subgroup of $G$ is normal. Since $7^1$ is the highest power of $7$ to divide $56$, every $P\in \text{Syl}_7(G)$ has order $7$. Taking another $Q\in \text{Syl}_7(G)$, we have that $P\cap Q=1$. Then the Sylow $7$-subgroups (excluding the identity) make up $6\times n_7$ distinct elements of $G$, all of which have order $7$. We have assumed that $n_p\not= 1$, so since $n_7\equiv 1 \mod 7$ and $1+6\times(7m+1)>56$ for any $m\geq 2$, we find that $n_7=8$.
That leaves us with $56-(8\times 6)=8$ elements with which we can form Sylow $2$-subgroups. Since $2^3=8$ is the highest power of $2$ dividing $56$, this is exactly enough for one Sylow $2$-subgroup, which must then be normal.
So that proves the implication $n_7\not= 1 \Rightarrow n_2= 1$, which we can extend to $(n_7=1)\vee (n_2=1)$. Thus $3$ is true if we take the "or" to be inclusive. (Of course, this implies statement $4$ even with a nontriviality condition.)
We still need to examine statements $1$ and $2$, however. Both statements are false, so we need to construct counterexamples for each. For statement $1$,
Assume that $n_7\not= 1$ and let $P\in \text{Syl}_7(G)$ and $Q\in \text{Syl}_2(G)$. Then $G=Q\rtimes P$ and $P$ acts on $Q$ by conjugation. The orbit of any element of $Q$ has size $7$ (since the semidirect product is nontrivial), so since $Q$ has size $8$, all nonidentity $q\in Q$ are in the same orbit. If we take $q_0$ to be the element of order $2$ in $Q$, we see that $o(q_0)=o(q^p)$. Thus $Q\cong \mathbb{Z_2}\times \mathbb{Z_2} \times \mathbb{Z_2}$.
Since $|GL_3(\mathbb{F}_2)|=(2^3-2^0)(2^3-2^1)(2^3-2^2)=7\times 6 \times 4$ is divisible by 7, $\text{Aut}(\mathbb{Z}_2^3)$ does indeed have an element of order $7$, so we can in fact form such a semidirect product. Thus there exists a group of order $56$ which does not have normal Sylow $7$-subgroups.
Constructing a counterexample for statement $2$ is a bit easier.
Let $P=\mathbb{Z}_8=\langle a \rangle$ and $Q=\mathbb{Z}_7$, so that $\text{Aut}(Q)\cong \mathbb{Z}_6=\langle \theta \rangle$. Let $G$ be the semidirect product $Q\rtimes_\phi P$ defined by $a\mapsto \theta^3$. Then $G$ is a group of order $56$ and $P$ is not normal in $G$.
In summary, to provide a complete answer to that question, you need Sylow theorems for statement $3$ and knowledge of nontrivial semidirect products to disprove statements $1$ and $2$.