Converting NFA to DFA

Question 1

If state Q2 gets an input of 'a' next states may be either Q1,Q2, 0r Q4.

In your NFA your get final state Q4

Its equivalent DFA is as below:

                 a-  
                 ||
                 ▼|
--►(Q0)---a---►((Q1))---b----►((Qf))  
                 ▲-----a--------|

Where Q1 and Q2 are final state.

And its Regular Expression is: a (a + ba)* (b + ε )

Where ε is null symbol (epsilon)

Question 2

We begin to convert NFA to DFA with identifying empty-input-closure sets (starting from here i will denote empty-input-closure by L-closure).
L(1)=(1,2) We can visit 2 from 1 on empty input.
L(2)=(2) There is no empty input edge out from 2.
L(3)=(3) There is no empty input edge out from 3.
L(4)=(1,2,4) We can visit 1 from 4 and 2 from 1.

If 2 gets an input of 'a' would it become (1,4) or (1,2,4) because of the empty string?

If 2 gets an input of 'a', it would become L(1)UL(4)=(1,2,4).

Since our start node is 1 in NFA, in DFA it will be L(1) which is (1,2).

T((1,2),a)=L(1)UL(3)UL(4)=(1,2,3,4)  
T((1,2),b)=F  
T((1,2,3,4),a)=L(1)UL(3)UL(4)=(1,2,3,4)  
T((1,2,3,4),b)=L(4)=(1,2,4)  
T((1,2,4),a)=L(1)UL(3)UL(4)=(1,2,3,4)  
T((1,2,4),b)=F

Since 4 is an accept node in NFA, in DFA nodes including 4 will be accept nodes, which are (1,2,3,4) and (1,2,4).