For (1), show that $[G,N]$ is a subset of both $N$ (using its normality) and $[G,G]$.
And for (2), use $[zN,gN]=N\iff [z,g]N=N\iff [z,g]\in N$ (put $\forall g\in G$ in front if it helps).
The subgroup generated by $\tau$ is cyclic of order $4$; the subgroup generated by $\sigma$ is cyclic of order $4$; you correctly calculate (since $\tau$ and $\sigma$ commute) that $\langle \sigma,\tau\rangle=\langle\sigma\rangle\langle\tau\rangle$, and that the order is $8$.
This is an abelian group. As such, it must be (abstractly) isomorphic to either $C_8$, $C_4\times C_2$, or $C_2\times C_2\times C_2$. The latter has no subgroups of order $4$, so that can't be it. The first has a unique subgroup of order $4$, so that can't be it, either. So that means that $G$ is abstractly isomorphic to $C_4\times C_2$. As such, it definitely has direct product decompositions.
Being abelian, every subgroup is normal. So you are just looking for subgroups with trivial intersection, one of order $4$ and one of order $2$ (those are really the only possibilities, because of the structure theorem for finite groups).
For instance, can we find a $2$-element subgroup $K$ such that $G=\langle \tau\rangle \times K$? Note that $(2,5)=\tau\sigma^{-1}\in G$. What is the intersection of $\langle (2,5)\rangle$ with $\langle \tau\rangle$? With $\langle \sigma\rangle$? Can we find other subgroups of order $4$ and decompositions?
If you find it difficult to it directly with the permutations, find an explicit isomorphism to $C_4\times C_2$ and work in the latter, then translating back to $G$.
Best Answer
You have a normal subgroup of order $5$. Your calculations are already sufficient to show that the elements of this group other than the identity don't commute with $\tau$, or indeed any of its powers. So $\tau, \tau^2, \tau^3$ are not in the centre. $1=\tau^0=\tau^4$ is of course in the centre.
Suppose we have an element $\rho$ which is in the centre, and therefore does commute with $\tau$. You have shown that $\rho = \sigma^k\tau^m$ so that $$\rho \tau =\sigma^k\tau^{m+1}$$ $$\tau\rho=\tau\sigma^k\tau^m=\sigma^{2k}\tau^{m+1}$$
For these to be equal you need $k=0$ - so the only possible central elements other than $1$ are the powers of $\tau$ which have already been excluded.