The idea of making DAG of a given expression is to construct corresponding syntax tree first and then we find the common occurence in the tree and join these common occurence to their common parent ; indicating that the same units are not taken repeatedly..This is shown as follow :

Hence number of internal nodes  =  3

Number of leaf node   =   1

Number of edges in DAG  =   6

In the given question , L_U is a set of strings which are accepted by the Universal Turing Machine..In order words , these are the strings in which the given Turing machine halts and each of the string of L_U has this property..In other words , L_U  represents a recursive language (or a recursive set)..

Now given reduction :   L  ≤  L_U

Now we know that for easy problems (or) languages [ REC (or) decidable , RE , P , NP ] , reduction works right to left i.e. if the class of the right part of the reduction is known , then we can deduce the left part also..But other way round , we cannot conclude anything..

As here , L_U is a recursive language , so due to the reduction : L  ≤  L_U  , L is recursive as well..

Hence option C) should be correct..

L1 = {0^n+m1^k+l | m = l, m, n, k, l >=1}
We can write it as
L1 = {0ⁿ0^m1^k1^l | m = l, m, n, k, l >=1},Since m and l does not depend upon n and k, we can fix its value as 1, and use n and k to take remaining 0's and 1's, so our language become..
L1 = {0ⁿ01^k1 |  n, k >=1}^,  Now we just need to check for any number of 0 followed by a 0, then any number of 1 followed by a 1, which is regular.

L2 = {0ⁿ(11)^m | m,n >=0}
Any number of zeros followed by even number of 1's, its regular.

According to me option D is correct, both L1 and L2 are regular.

$8\\ 8\times1-2=6\\ 6 \times2-3=9\\ 9 \times 3 - 4=23\\ 23 \times 4-5=87\\ 87 \times 5-6=429$

Hence, Option(D)429 is the correct choice.

Here we are given that the order of the group is 10 and a is the generator..

Hence  a¹⁰  =  e  where e is the identity element but we dont know about a⁸..Hence we have to write it in terms of a¹⁰ as that is a known result.

Let we have   y  =  a⁴⁰   which can be written as :    a⁴⁰  =   (a¹⁰)⁴   =  e⁴   =   e    ...(1) 

                                                            Also ,        a⁴⁰   =  (a⁸)⁵     =  e     [ Follows from (1) ]

Hence it means that  a⁸ is repeated 5 times in order to get the identity element e which is least number of times to do so.

Also from the corollary of the Lagranges' Theorem , 

Order of an element  divides the order of the order of the group (The actual theorem is order of a subgroup divides the order of the group)

Here 5 divides the order of group i,e, 10..

Hence order of  a⁸   =   5