Group theory conjectures

You keep talking about "homomorphisms" which turn out not to be homomorphisms. Your first $\phi_2$ has $\phi_2(1)=2$, $\phi_2(2)=4$ and $\phi_2(3)=3$. If $\phi_2$ were a homomorphism, then we would have $$\phi_2(2)=\phi_2(1+1)=\phi_2(1)+\phi_2(1)=2+2=4,$$ (good), $$\phi_2(3)=\phi_2(2+1)=\phi_2(2)+\phi_2(1)=4+2=1$$ (as we are in $\Bbb Z_5$), which doesn't agree with your original $\phi_2$, so that isn't a homomorphism. But it shows us how to create one; next, $$\phi_2(4)=\phi_2(3+1)=\phi_2(3)+\phi_2(1)=1+2=3,$$ $$\phi_2(5)=\phi_2(4+1)=\phi_2(4)+\phi_2(1)=3+2=0,$$ etc.

The principle is that when we know what $\phi(1)$ is, we can determine what the homomorphism $\phi$ is. So, if $\phi(1)=4$, then $\phi(2)=4+4=3$. So your $\phi_4$ is not a homomorphism.

Anyway, for each homomorphism $\phi$, $\phi(5)=\phi(1)+\cdots+\phi(1)$ (five summands) which equals zero. Then $\phi(10)=\phi(5)+\phi(5)=0+0=0$ etc. The kernel contains $\{0,5,10,\ldots,25\}$. But the kernel of a surjective homomorphism has order $30/5$, so that must actually be the kernel.

These are true and pretty straightforward. I didn't read through all the examples because I think you're making this too complicated for yourself. In the first bullet, note that the existence of the onto homomorphism forces $m|n$. Since an onto homomorphism must take a cyclic generator (which you can take to be $1$) to a cyclic generator, precisely the $m^{th}$ multiples of the generator will be in the kernel.

For the second item, note that the homomorphism can only take a cyclic generator to the subgroup of order $(m,n)$ in the target. It can take it to any element of that subgroup, and there are $(m,n)$ ways to do this.

The main properties used here are the a homomorphism from a cyclic group to any other group is completely determined by the target of any generator, and the only restriction on the target of that generator is that its order must divide the order of the generator (and this restriction only applies if the source cyclic group is finite).