Solution to part (b) is extremely confusing.
First, define an addable integer to be one which is attainable with the last operation being addition. Similarly, define a multiplicable integer to be one which is attainable with the last operation being multiplication. For the sake of convenience, let $1$ be both addable and multiplicable.
a.) Clearly when $n$ is even we are done. Otherwise, consider the sequence $a_1=n+4$, and defined recursively as $a_{i+1}=2a_i+n$ for $a_i\not \equiv 1\pmod n$ and $a_{i+1}=4a_i+n+4$ for $a_i\equiv 1\pmod n$. It's important to note that we always have $a_i\equiv 1\pmod 2$ and $a_i\not \equiv 0,2\pmod n$. I claim that every term in the sequence is unattainable. It's easy to see that $a_1=n+4$ is unattainable. Now, suppose that $a_i$ is unattainable and $a_i\not\equiv 1\pmod n$. Then note that if $a_{i+1}$ is attainable then the last two operations must be $\times 2$ followed by $+n$. This implies that $a_i$ is addable, a contradiction. Otherwise, suppose that $a_i\equiv 1\pmod n$. Then if $a_{i+1}$ is attainable then $2a_i+2$ is addable. This implies that $a_i$ is addable, a contradiction.
b. We attempt to find all numbers which aren't addable. If we work backwards, we see that we can always find a path which leads back to either $1$ or $2$ (the proof being one by contradiction). For example, we have $$20\rightarrow 18\rightarrow 9\rightarrow 6\rightarrow 3\rightarrow 1$$and $$34\rightarrow 32\rightarrow 16\rightarrow 14\rightarrow 7\rightarrow 4\rightarrow 2.$$Ignore every other step; they are irrelevant to the solution. Now, if a number isn't addable, then any paths leading from it must end at $2$. If this occurs, then the second to last number in the sequence is either $6$, $7$, $8$, or $9$. Checking, we find that if the number is $6$, $8$, or $9$, there is some alternate path which leads back to $1$. Now, consider the sequence $b_1=7$, and $b_{i+1}=2b_i+2$ for all $i\ge 1$. Consider our sequence after $k$ steps. Then I claim that the last numbers in the path must be $b_k,b_{k-1},\dots,b_1,2$. We have already shown the base case of $k=1$. Otherwise, note that the $k+1$th to last number in the path must be $2b_{k-1}+2$, $2b_{k-1}+3$, $3b_{k-1}+2$, or $3b_{k-1}+3$. If it's one of the last three, we can just find a new path where the $k$th to last number isn't $b_{k-1}$ by selecting different operations for that step. However, if $2b_{k-1}+2$ is the $k+1$th to last number, we're forced to have the $k$th to last number be $b_{k-1}$.
This is enough to imply that all numbers in this path, if we continued infinitely, aren't addable, while all numbers not in this path are. For all even numbers in the path, we can note that they are multiplicable since we can just divide by $2$ and obtain an addable number. The only odd number in the path is $7$. This is unattainable by finite case check. We are done. $\blacksquare$