First of all, statement $2$ is false :
Since the sum of the digits of $N$ is $43\equiv 7\pmod9$, we know that $N\equiv 7\pmod9$. However, since there is no $m\in\mathbb N$ such that $m^3\equiv 7\pmod9$, we know that statement $2$ is false.
As a result, we have to find $N$ which satisfies both statement $1$ and $3$. Here, note that
$$m^2\equiv 7\pmod9\iff m\equiv 4,5\pmod9.$$
This fact will enable you to eliminate many cases.
So, we have $66$ cases to check if the sum of the digits of $m^2$ is $43$ :
$$m=418+9t, 419+9t\ \ (t=0,1,\cdots,32).$$
As you have observed, you cannot have any number ending with a $0$, so your first two digits are fixed.
Division tests for numbers ending with $1,2,5,6$ add no information.
From the division test for $3$, you know that the first two digits must sum to a multiple of $3$.
From the division test for $7$, you know that the number obtained by truncating the last digit and subtracting away twice the last digit from it will give a multiple of $7$. This means that the first two digits have to also be a multiple of $7$.
From these two conditions, the first two digits must be a multiple of $21$, which means they can be $21,42,63,84$.
From the division test for $4$, you know the last two digits must form a number divisible by $4$.
The only one that fits this condition is the first two digits $84$.
So the smallest number is $841$, digit sum $13$.
For rigour, you need to exclude the possibility of the sequence starting with first digit $2$. If this were the case, the final number would end with $9$, and the division test for $9$ demands the sum of the first two digits should also be a multiple of $9$. But this would mean that $63$ is the only possibility, which would conflict with the rule for the last digit $4$ ($34$ is not a multiple of $4$). So we have verified that the above is the only sequence of numbers that meets the criteria.
Best Answer
We compute all the possible residues $n(n+1)\bmod10$, which are $[0, 2, 6, 2, 0, 0, 2, 6, 2, 0]$. If the last digit of $\frac{n(n+1)}2$ was a $9$, the last digit of $n(n+1)$ would be $8$, but since $8$ does not appear as a possible value of $n(n+1)\bmod10$, $9$ cannot be the last digit of $\frac{n(n+1)}2$.