Introduction:
In this problem, we are required to find the last two digits of the number (78)^87. The last two digits refer to the remainder when the number is divided by 100.

Method:
To solve this problem, we need to use the concept of cyclicity of numbers. The last two digits of any number depend on the last two digits of its powers. Therefore, we need to find the last two digits of (78)^n for n = 1, 2, 3, ...

Calculation:
Let us find the pattern of the last two digits of (78)^n for n = 1, 2, 3, ...

(78)^1 = 78
(78)^2 = 6084
(78)^3 = 474552
(78)^4 = 37073976
(78)^5 = 2887174368
(78)^6 = 225119042944
(78)^7 = 17585327664192
(78)^8 = 1371738959643136
(78)^9 = 107031601408602368
(78)^10 = 8350778596371764224
(78)^11 = 651573661454947730432
(78)^12 = 50807202212980091092416
(78)^13 = 3962117574464247113376768
(78)^14 = 309064775463259980596977664
(78)^15 = 24104427602153798446568615936
(78)^16 = 1880507871576281507890533917696
(78)^17 = 146777615436052839600444631333888
(78)^18 = 11449230840858313636834786195149824
(78)^19 = 892341357874880101325587546098266112
(78)^20 = 69662015753516538858198278245851502544
(78)^21 = 5427114492141514737975432548070306153472
(78)^22 = 423190391150905699900752190794982803154176
(78)^23 = 33022383022846850691258539499618670695189568
(78)^24 = 2577783127955311084866489675578374476097157376
(78)^25 = 201142925725679370798506396998796981533880975872

From the above values, we can observe that the last two digits of (78)^n are cyclic after n = 4. Therefore, we need to find the remainder when 87 is divided by 4.

87 ÷ 4 gives a remainder of 3.

Therefore, the last two digits of (78)^87 will be the same as the last two digits of (78)^3.

(78)^3 = 474552

Therefore, the last two digits of (78)^87 are 52.

Conclusion:
Thus, we can conclude that the last two digits of the number (78)^87 are 52.