Hints: There are $\binom{10}{4}$ ways to choose a "hand" of $4$ cards from the $10$. All hands are equally likely. So to answer the three questions, all we need to do is to count the hands that qualify, and divide each count by $\binom{10}{4}$. The numbers are small, we could count by making careful lists.
$1.$ How many hands have $4$ consecutive cards? Surely we can list them all, without any theory. There is $\{5,6,7,8\}$, and $\{6,7,8,9\}$, and $\{7,8,9,10\}$.
$2.$ How many aceless hands are there? We need to choose $4$ cards from the $6$ non-aces.
$3.$ We can choose the $2$ Aces in $\binom{4}{2}$ ways. For each of these ways, the two consecutives can be chosen in $5$ ways.