a. This is just shifting the index of summation:
[tex]\displaystyle \sum_{j=0}^4 d_{j+1} = d_{0+1}+d_{1+1}+\cdots+d_{4+1} \\ = d_{2-1}+d_{3-1}+\cdots+d_{6-1} \\ = \sum_{j=2}^6 d_{j-1}[/tex]
and just swap j with k, since you can freely use any symbol for the index.
That is, the identity relies on simple integer arithmetic: 0 + 1 = 1 = 2 - 1, and 1 + 1 = 2 = 3 - 1, and so on.
b. This is something of a generalization of the identity in part a.
[tex]\displaystyle \sum_{j=1}^n a_jb_{m+j} = a_1b_{m+1} + a_2b_{m+2} + \cdots + a_nb_{m+n} \\ = a_{(m+1)-m}b_{m+1} + a_{(m+2)-m}b_{m+2} + \cdots + a_{(m+n)-m}b_{m+n} \\ = \sum_{j=m+1}^{m+n} a_{j-m}b_j[/tex]
c. Distribute the summation to get
[tex]\displaystyle \sum_{j=1}^m (6d_j+1) = \sum_{j=1}^m 6d_j + \sum_{j=1}^m 1[/tex]
The second sum here is just adding together m copies of 1,
[tex]\displaystyle \sum_{j=1}^m 1 = \underbrace{1+1+\cdots+1}_{m\,\rm times} = m[/tex]
so
[tex]\displaystyle \sum_{j=1}^m (6d_j+1) = \sum_{j=1}^m 6d_j + m[/tex]
d. The sum as given is generally not true:
[tex]\displaystyle \sum_{n=1}^k a_k = a_1+a_2+\cdots+a_k[/tex]
while (generalizing the identity from part c)
[tex]\displaystyle \sum_{m=2}^{k+1} (a_m-1) = (a_2+a_3+\cdots+a_k+a_{k+1})-k[/tex]
and these expressions are equal only if [tex]a_1=a_{k+1}-k[/tex].
