For the first problem, anything that starts with an a, or just b, or anything that starts with a b then not an a:
a(a + b)* + b + bb(a + b)*
In a systematic way:
If we expand all possible strings, making the first two symbols explicit, we have
a + b + aa(a + b)* + ab(a + b)* + ba(a + b)* + bb(a + b)*
From these, we discard those beginning with ba, leaving
a + b + aa(a + b)* + ab(a + b)* + bb(a + b)*
Optionally we can compress the three terms beginning in a by
a + aa(a + b)* + ab(a + b)* = a + a(a + b)(a + b)* = a(a + b)*
and a final grouping is possible
a(a + b)* + b + bb(a + b)* = (a + bb)(a + b)* + b
The solution in the OP isn't valid as it does allow ba (choice b in the parenthesis, then a* once).
For the second problem, let us consider all strings with multiples of three b's
(bbb)*
and let us insert abritrary strings of a wherever possible: we get
a* (a* b a* b a* b a*)* a*
This solution is valid. Anyway, it leaves multiples possibilities for the decomposition of the initial and final strings of a, because of sequences a*a*. These can be avoided by the mergings a*a* = a*, which give the unambiguous
a* (b a* b a* b a*)*
The solution in the OP isn't valid as is does not represent the strings without any b, though they should be accepted.
Best Answer
One very straightforward approach to (a) is to begin by listing the $8$ binary strings of length $3$. Every string whose length is a multiple of $3$ must be formed by concatenating those strings. I’ll illustrate with strings of even length: any string of even length can be segmented into two-character chunks, each of which must be $00,01,10$, or $11$, so the binary strings of even length are described by the regular expression $(00+01+10+11)^*$. (You may use some other symbol than $+$ to indicate alternatives; common alternatives are $\mid$, $\lor$, and $\cup$.)
Problem (b) is a little harder. The condition says that if a string contains a $1$ at all, that $1$ must either be the last symbol in the string or be followed immediately by a $0$. Ignore for a moment the exceptional case of a final $1$; if we just require that every $1$ be followed immediately by a $0$, we’re allowing precisely those strings that can be built by concatenating $0$’s and $10$’s, i.e., the strings described by the regular expression $(0+10)^*$. We’d like to have all of those strings, but also all strings that consist of one of those followed by a single $1$; how can you modify or extend the regular expression to include the latter type as well as the former?