Create vector $$ such that $^x$, for a given vector $x$ in each part.
1) $^$ extracts (is equal to) the $7$th entry of the $10$-vector $$
$a = [0, 0, 0, 0, 0, 0, 1, 0, 0, 0]$
2) $^$ is the weighted average of a $3$-vector $$, assigning weights $0.3$ to the first component, $0.4$ to the second, and $0.3$ to the third.
$a = [0.3/3, 0.4/3, 0.3/3]$
3) $^$ (with $$ a $22$-vector) is the sum of $_$ for $$ a multiple of $4$, minus the sum of $_$ for $$ a multiple of $7$.
4) $^$ (with $$ an $11$-vector) is the average of the middle three entries of $$, i.e., entries $5$ to $7$.
I'm having trouble with 3 and 4.
2) Your answer here isn't correct: using the given weights already creates the desired weighted average, so dividing by $3$ on top of that is unnecessary and wrong.
3) Same logic as you used in the first problem. Here you want all elements whose index (position) is a multiple of $4$ to be taken with a coefficient of $1$ thus adding them together, and at the same time you want all elements whose index (position) is a multiple of $7$ to be taken with a coefficient of $-1$ thus adding subtracting all of them.
4) You need to take the average of the three entries: the $5$th, the $6$th, and the $7$th ones, so you here (unlike in the second question) you need to multiply each of them by $1/3$, while you "kill off" all the other entries with zeroes like you did (correctly!) in the first problem.