Your strategy is sound, but your answer is indeed incorrect.
Case 1: Sione receives two cookies.
Let $x_i$, $1 \leq i \leq 4$, be the number of cookies distributed to the $i$th friend. Let $x_4$ denote the number of cookies given to Sione. Since a total of ten cookies are distributed and Sione receives two,
\begin{align*}
x_1 + x_2 + x_3 + 2 & = 10\\
x_1 + x_2 + x_3 & = 8 \tag{1}
\end{align*}
Since each friend receives at least one cookie, equation 1 is an equation in the nonnegative integers. A particular solution of equation 1 corresponds to the placement of two addition signs in the seven spaces between successive ones in a row of eight ones.
$$1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1$$
For instance, if we place an addition sign in the third and seventh boxes, we obtain
$$1 1 1 + 1 1 1 1 + 1$$
which corresponds to the solution $x_1 = 3$, $x_2 = 4$, and $x_3 = 1$. The number of such solutions is the number of ways we can select two of the seven spaces in which to place an addition sign, which is
$$\binom{7}{2}$$
A particular solution of the equation
$$x_1 + x_2 + x_3 + \cdots + x_n = k$$
in the positive integers corresponds to the placement of $n - 1$ addition signs in the $k - 1$ spaces between successive ones in a row of $k$ ones. Therefore, the number of such solutions is
$$\binom{k - 1}{n - 1}$$
since we must choose which $n - 1$ of those $k - 1$ spaces will receive an addition sign.
Case 2: Sione receives four cookies.
Let the variables be assigned as above. Since Sione receives four cookies,
\begin{align*}
x_1 + x_2 + x_3 + 4 & = 10\\
x_1 + x_2 + x_3 & = 6 \tag{2}
\end{align*}
Since each friend receives at least one cookie, equation 2 is an equation in the positive integers with
$$\binom{6 - 1}{3 - 1} = \binom{5}{2}$$
solutions.
Case 3: Sione receives six cookies.
Let the variables be assigned as above. Since Sione receives six cookies,
\begin{align*}
x_1 + x_2 + x_3 + 6 & = 10\\
x_1 + x_2 + x_3 & = 4 \tag{3}
\end{align*}
Since each friend receives at least one cookie, equation 3 is an equation in the positive integers with
$$\binom{4 - 1}{3 - 1} = \binom{3}{2}$$
solutions.
Total: Since these cases are mutually exclusive and exhaustive, you can distribute the cookies in
$$\binom{7}{2} + \binom{5}{2} + \binom{3}{2} = 21 + 10 + 3 = 34$$
ways.
For 1, you just choose the seven children who will get gifts, so $10\choose 7$ ways
Your answer to 1 is the correct approach for 2
Best Answer
If $10$ gifts are distributed among three friends and each friend receives at least three gifts, then one person receives four gifts and the others each receive three. There are three ways to select the person who receives four gifts, $\binom{10}{4}$ ways to select the gifts that person receives, $\binom{6}{3}$ ways to select the gifts the younger of the two remaining friends receives, and the other friend must receive all of the remaining three gifts. Hence, there are $$\binom{3}{1}\binom{10}{4}\binom{6}{3}\binom{3}{3}$$ admissible ways to distribute the gifts.
Note: Your method counts each distribution four times, once for each of the $\binom{4}{3}$ ways of designating three of the four gifts the person who receives four gifts as the three gifts that person receives.