Thanks for your patience Amber.

In 7.13, we just need to list out a few terms of the recursive sequence. For example, we know $a_0 = 2$, so $$a_1 = 2a_0 + 0 = 2\times 2 + 0 = 4,$$ then $$a_2 = 2a_1 + 1 = 2\times 4 + 1 = 9,$$ and $$a_3 = 2a_2 + 2 = 2\times 9 + 2 = 20.$$ Hope this helps you see the pattern and you can get $a_4$.

In 7.26b, the fifth term and eighth term are $3$ terms apart. Therefore if $r$ is the common ratio, the sixth term is $8r$, then the seventh is $8r^2$, and the eighth is $8r^3$. Using this fact, can we somehow find the common ratio? Hint:  How do we "undo" cubing a number?

In 7.27, look back at the trick used in 7.7(a) and (b). The same idea (writing the sum forwards and backwards) should work here as well.

in 7.29, trying to find a pattern is the best way to try to solve the problem. For example, $1$ can just be written as $1$, so there is $1$ way. For $2$ we only have $1+1$, so there is also only $1$ way for $2$. For $3$ we have $$3 \text{ or } 1 + 1 + 1,$$ so there are $2$ ways for $3$. For $4$ we have $$3 + 1 \text{ or } 1+3 \text{ or } 1 + 1 + 1 + 1$$ so there are $3$ ways for $4$. Have we seen a sequence that starts $1, 1, 2, 3$ before that could be our pattern?

For 7.30, you can ignore the other wording on this page (click here: https://www.geeksforgeeks.org/perfect-binary-tree-specific-level-order-traversal/ ) but this gives a good picture of a larger binary tree.

Lastly, for 7.10b, remember it is $2^n - 2$, with an exponent. In (a) we have that for $1$, $2$, $3$, and $4$ friends there are $0$, $2$, $6$, and $14$ ways to invite them. This matches the formula $2^n - 2$ as $2^1 - 2 = 2 - 2 = 0$, $2^2 - 2 = 4-2 = 2$, $2^3 - 2 = 8 - 2 = 6$, and $2^4-2 = 16-2 = 14$.