Your questions:
1) This is not enough: it must be also that both sbgps. are normal in G
2) Because they're abelian, too.
3) Lots more, as anyone else...but not for this particular question, imo.
$\mathbb{Z}_5\times \mathbb{Z}_5$ can be considered as a vector space over field $\mathbb{Z}_5$ (the addition is same as addition in abelian group modulo $5$, and scalar multiplication actually comes from addition: $v+v=2.v$, $v+v+v=3.v$, $\cdots$)
Thus any automorphism of the group $\mathbb{Z}_5\times \mathbb{Z}_5$ is also an automorphism of the vector space $\mathbb{Z}_5\times \mathbb{Z}_5$ and conversely (because of remark on scalar multiplication made earlier).
It is well known, what is the automorphism group of $\mathbb{Z}_5\times \mathbb{Z}_5$? It is
$$
GL(2,5)=
\begin{Bmatrix}
\begin{bmatrix}
a & b\\
c & d
\end{bmatrix}\colon a,b,c,d\in\mathbb{Z}_5, ad-bc\neq 0
\end{Bmatrix}.
$$
To get an element of order $3$ in $Aut(\mathbb{Z}_5\times \mathbb{Z}_5)$, we have to find a non-identity matrix $A$ such that $A^3=I$. Thus, $A$ satisfies polynomial $x^3-1=(x-1)(x^2+x+1)$. The quadratic factor has no root in $\mathbb{Z}_5$ (check) so it is irreducible. Further, $A\neq I$, hence the minimal polynomial of $A$ should divide quadratic factor $x^2+x+1$. But the minimal polynomial of $A$ has degree $\leq 2$ (size of $A$), so $x^2+x+1$ is the minimal (hence characteristic polynomial of $A$). Can we find a matrix $A$ explicitly with such characteristic polynomial? Yes; companion matrix
$$
A=
\begin{bmatrix}
0 & -1\\
1 & -1
\end{bmatrix}.
$$
Now we come to construction of group. Let $\mathbb{Z}_5\times \mathbb{Z}_5=\langle x,y\rangle$ with $x=\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $y=\begin{bmatrix} 0 \\ 1 \end{bmatrix}$. Then $Ax=y$ and $Ay=-x-y$ (check).
Consider $\mathbb{Z}_5\times \mathbb{Z}_5=\langle x,y\rangle$ multiplicatively, $t$ be element of order $3$ in its automorphism group, and action of $t$ on this group is given by (write above action of $A$ as action of $t$, with multiplicative operation):
$$txt^{-1}=y \,\,\,\, tyt^{-1}=x^{-1}y^{-1}.$$
Thus
$$G=\langle x,y,t\colon x^5=y^5=1, xy=yx, t^3=1, txt^{-1}=y, tyt^{-1}=x^{-1}y^{-1}\rangle$$
is a group of order $75$, it is non-abelian.
Best Answer
(Note: I just finished teaching a lecture on semi-direct products and their constructions, so my mind went there, especially as I know the structure of the automorphism groups at issue well; but as Steve D points out in comments, you don't even need the automorphism groups and there is a simpler solution that only uses the Sylow theorems, and which I have added below under the horizontal line.)
The only additional thing you need is to know the automorphism groups of the nonabelian groups of order $27$, and of the abelian groups of order $27$.
Let $G$ be a group of order $27p$ with $p$ a prime, $p\gt 3$ and $3\nmid p-1$. The number of $3$-Sylow subgroups of $G$ is congruent to $1$ modulo $3$ and divides $27p$, hence is either $1$ or $p$. But if $3\nmid p-1$, then it cannot be $p$. So the $3$-Sylow subgroup of $G$ is normal. Call it $N$.
Let $C_p$ be a subgroup of order $p$ of $G$ (necessarily cyclic). Then $G$ is a semidirect product of $N$ by $C_p$, since $G=NC_p$ and $N\triangleleft G$. We just need to see what kind of actions $C_p$ can have on $N$.
We have $5$ cases:
$N$ is cyclic of order $27$. The automorphism group of $N$ is cyclic, of order $\phi(27) = 18 = 2\times 3^2$, so the action of $C$ on $N$ is trivial, and we have a direct product, which is cyclic of order $27p$.
$N$ is isomorphic to $C_9\times C_3$, where $C_n$ is the cyclic group of order $n$. As shown here, the automorphism group here has order $\phi(9)\phi(3)3^2 = 4\times 3^4$, and again that means that any map $C_9\to\mathrm{Aut}(C_9\times C_3)$ is trivial, so we have a direct product, which is isomorphic to $C_3\times C_{9p}$.
$N$ is elementary abelian. Then the automorphism group of $N$ is isomorphic to $\mathsf{GL}(3,3)$, which has order $(3^3-1)(3^3-3)(3^3-9) = (26)(24)(18)$. The prime factors are $2$, $3$, and $13$, so you could have a nontrivial semidirect product $N\rtimes C_{p}$ if $p=13$... but in that case $3\mid 13-1$, which we have excluded. For any other prime $p\gt 3$ the action is trivial, so again we get an abelian group, this time isomorphic to $C_3\times C_3\times C_{3p}$.
$N$ is the Heisenberg group of order $27$. As noted here, the automorphism group is isomorphic to $$\mathsf{AGL}(2,3) = \left\{ \begin{pmatrix}a & b& e\\ c& d& f\\ 0 & 0 & 1\end{pmatrix} : a,b,c,d,e,f \in \mathbb{Z}/3\mathbb{Z},\; ad-bc \neq 0 \right\}.$$ The order of the group is $9$ times the number of choices for $a,b,c,d$. This is equal to the number of ordered pairs of linearly independent vectors in $\mathbb{F}_3^2$, which is $(9-1)(9-3)=2^4\times 3$; so any homomorphism from $C_p$ to $\mathsf{AGL}(2,3)$ is trivial, hence the action of $C_p$ on $N$ is trivial, and $G\cong N\times C_p$.
$N$ is the nonabelian group of order $27$ that has an element of order $9$. Again in Jack Schmidt's post quoted above he notes the automorphism group can be identified with $$\left\{ \begin{pmatrix}a & b& 0\\ 0& 1& 0\\ c & d & 1\end{pmatrix} : a,b,c,d \in \mathbb{Z}/3\mathbb{Z},\; a ≠ 0 \right\}.$$ This group has order $2\times 3^3$, so any morphism from $C_p$ to the automorphism group of $N$ is trivial and we get $G\cong N\times C_p$.
So the only groups of order $27p$ with $p$ primes, $p\gt 3$, $3\nmid p-1$ are abelian, or of the form $N\times C_p$ where $C_p$ is cyclic of order $p$ and $N$ is nonabelian of order $27$. If you drop the restriction on $3\nmid p-1$ but require the $3$-Sylow to be normal, then you get another nonabelian group by letting $C_{13}$ act on the elementary abelian group of order $27$.
A simpler solution uses only the Sylow theorems, as pointed out by Steve D below. We already know there is a single $3$-Sylow subgroup, so to prove that $G\cong N\times C_p$ it is enough to show there is a single $p$-Sylow subgroup. The number of $p$-Sylow subgroups must divide $27p$ and be congruent to $1$ modulo $p$, so it must divide $27$. The divisors of $27$ greater than $1$ are $3$, $9$, and $27$. The only prime such that $3\equiv 1\pmod{p}$ or $9\equiv 1\pmod{p}$ is $p=2$, which is excluded for being smaller than $3$. The primes for which $27\equiv 1\pmod{p}$ are $2$ and $13$, and $2$ is excluded for being smaller than $3$, and $13$ because $3\mid 13-1$. Thus, there must be a unique Sylow $p$-subgroup as well, so $G\cong N\times C_p$. The abelian $N$ give abelian groups, the nonabelian groups have the desired form.