Here is a slightly indirect way of obtaining the variance:
Let $X_k$ be the number on the $k$th ticket, $k=1,2,\ldots,m$.
So we have a uniform distribution for the $X_k$'s, namely
$$ P(X_k=j)=\begin{cases}\frac{1}{n}&,\text{ if }j=1,2,\cdots,n\\\\\,0&,\text{ otherwise }\end{cases}$$
So,
\begin{align}
\operatorname{Var}(X_k)&=E(X_k^2)-(E(X_k))^2
\\\\&=\frac{n^2-1}{12}=\sigma^2\,,\text{ say }
\end{align}
If the correlation between $X_i$ and $X_j$ $\,(i\ne j)$ be $\rho$, then $$\rho=\dfrac{\text{Cov}(X_i,X_j)}{\sigma^2}$$
You are looking for \begin{align}\operatorname{Var}(X)&=\operatorname{Var}\left(\sum_{k=1}^m X_k\right)\\&=\sum_{k=1}^m \operatorname{Var}(X_k)+2\sum_{i<j}\text{Cov}(X_i,X_j)\\&=m\sigma^2+2\binom{m}{2}\rho\sigma^2
\\&=m\sigma^2(1+(m-1)\rho)\tag{1}\end{align}
Now note that the joint distribution of $(X_i,X_j)\,,i\ne j$ is independent of $m$.
So we see that
\begin{align}
\operatorname{Var}\left(\sum_{k=1}^{\color{red}{n}}X_k\right)&=\operatorname{Var}(\text{constant})=0
\\&\implies\color{red}{n}\sigma^2(1+(\color{red}{n}-1)\rho)=0
\\&\implies\rho=\frac{1}{1-n}
\end{align}
Substituting this value of $\rho$ and the value of $\sigma^2$ in $(1)$, we finally get the variance of $X$ as
$$\operatorname{Var}(X)=\frac{m(n+1)(n-m)}{12}$$
For the first one: Let $P_n$ be the probability that the largest draw is $≤n$. Of course, the probability that any given draw is $≤n$ is $\frac n{15}$, thus $$P_n=\left( \frac n{15} \right)^7$$
Of course, the probability that the max is at exactly $n$ is then $P_n-P_{n-1}$, so in this case the answer you want is $$\left( \frac 9{15} \right)^7-\left( \frac 8{15} \right)^7$$
Which is not the answer you report. Your answer would be correct if you are just requiring that the maximum draw be $≤9$ but that isn't the question. Perhaps you mistyped the first question?
The second problem can be answered along the same lines.
Best Answer
Let $X_i$ be the value of the $i^{th}$ ticket drawn. Let $X$ denote the largest value drawn, and notice that $X = \max{ \{ X_1, ..., X_n \}}$. Let's try to find the pmf of $X$. This is actually easiest if we use CDFs.
$F_X(k)=P(X \leq k)=P(\max{ \{ X_1, ..., X_n \}} \leq k) = P((X_1 \leq k) \cap ... \cap (X_n \leq k))$.
The second equality holds because we need each of the $X_i$ to be less or equal to $k$ if we want the maximum to be less or equal to $k$.
Now notice that we are sampling the tickets with replacement, so each draw is independent. This allows us to say
$P((X_1 \leq k) \cap ... \cap (X_n \leq k)) = P(X_1 \leq k)\cdot \cdot \cdot P(X_n \leq k) = P(X_i \leq k)^n$.
Again, since we are sampling with replacement, he last equality holds because each draw has the same distribution. Now we know that each $X_i$ is a discrete uniform distribution on the set $ \{ 1, 2, ... N \}$, so, putting it all together, what we end up with is
$F_X(k) = P(X_i = k)^n = (k/N)^n$.
From this we can easily find $p_X(k)$, and then $E(X) = \sum_{k=1}^{N} kp_X(k)$