a. the bride must be in the picture

Let us divide this problem into cases.

Code:

BXXXXX
XBXXXX
XXBXXX
XXXBXX
XXXXBX
XXXXXB

Where "X" stands for **any** other person. And "B" stands for bride. We will find the number of such cases and then add all of them up together.

In the first case there are $\displaystyle (1)(9)(8)(7)(6)(5) = 15120$. Similarly in each case we have the same number of arrangements. So in total we have $\displaystyle 6\cdot 15120 = 90720$.

b. exactly one of the bride and the groom is in the picture

Again divide this into cases. Case 1 is that bride is in picture and groom is not. Case 2 is that bride is not in picture and groom is. Then add those results together. Each subcase is similar to the problem just did above.