If in a particular year 'X' there are 53 Sundays then how many Sundays will be there in a period of four years X to X + 3 year.

In a year we have 53 Sundays only when year start with Sunday (For non leap year) and either with Saturday of Sunday (For leap year).
From zeller's formula 1^st January 2001 was Monday
From number of odd days 1^st January 2006 will start with Sunday so it had 53 Sunday
Similar 1^stJanuary 2009 will start with Thursday.

Number of odd days between 2013 and 2015 is 2, so we have to find the year which will have 2 odd days between 1977 and required year.
Consider options one by one -
(a) Number of odd days between 1977 and 1981 is 1 + 1 + 1 + 2 = 5 odd days
(b) Number of odd days between 1977 and 1985 is 1 + 1 + 1 + 2 + 1 + 1 + 1 + 2 = 10 odd days or 3 odd days
(c) Number of odd days between 1977 and 1990 is 1 + 1 + 1 + 2 + 1 + 1 + 1 + 2 + 1 + 1 + 1 + 2 + 1 = 16 odd days or 2 odd days, hence calendar of 1990 is the answer.

Calendar of 2 years is similar if number of odd days between these two years is zero. Consider option one by one
(a) Between 1940 to 1946 we have two leap years 1940 and 1944 so number of odd days is 2 + 1 + 1 + 1 + 2 + 1 = 8 or 1 hence calendar of these two years is not similar.
(b) Between 1977 to 1982 we have one leap years 1980 so number of odd days is 1 + 1 + 1 + 1 + 2 + 1 = 6 calendar of these two years is not similar.
(c) Between 1912 to 1916 we have one leap years 1912 so number of odd days is 2 + 1 + 1 + 1 = 5 hence calendar of these two years is not similar.

In a period of 100 years there are 23 or 24 leap years (as for century year it might be or might not be a leap year, as 1900 was not a leap year)
Number of days in a period of 100 years is 365 x 100 + 23 or 365 x 100 + 24.
If century year is not a leap year then number of days = 365 x 100 + 23, and number of weeks is 5217 and 4 odd and for leap year it will be 5217 weeks and 5 odd days, hence number of Sundays is either 5217 or 5218.

Lets find out 1^st April from Zeller's Formula:
f = k + [13 x m - 1 / 5] + D + [D/4] + [C/4] - 2 x C.
In this case k = 1 (since 1^st April)
Month m = 2 (As march = 1, April = 2)
D is the last two digit of year here D = 01 (As year is 2001)
C is 1^st two digit of century here C = 20 (As year is 2001)
f = 1 + [13 x 2 - 1 / 5] + 01 +[01/4] + [20/4] -2 x 20
f = 1 + [25/5] + 01 + [0.5] + [5] - 40.
f = 1 + 5 + 01 + 0 + 5 - 40 = -28
This - ve value of f can be made positive by adding multiple of 7
So f = - 28 + 28 = 0
So number of odd days is 0,
So 1^st April 2001 is Sunday,
So 1^st Friday is on 6^th April , so next Fridays is 13^th, 20^th, 27^th April.

If month ends with Thursday then next month will start with Friday and it may have 5 Friday otherwise it may have 4 Fridays.

Counting the number of days after 3rd November. 1994 we have :
Number of Days in November = 27 or 6 odd days.
Number of Days in December = 31 or 3 odd days.
Number of Days in January = 31 or 3 odd days.
Number of Days in February = 28 or 0 odd days.
Number of Days in March = 20 or 6 odd days.
So total number of odd days = 6 + 3 + 3 + 0 + 6 = 18 when divided by 7 gives remainder 4.
Number of odd days = 4
The day on 3rd November, 1994 is (7 - 4) days beyond the day on 20th March, 1995.
So, the required day is Thursday.

94 when divided by 7 gives remainder 3 hence today it must be Thursday, after 3 more we will get a Sunday, next Sunday will be after 3 + 7 = 10 days and so on so we will get Sunday after a day 7K + 3 days.

100 years contain 5 odd days.
? Last day of 1^st century is Friday.
200 years contain (5 x 2) = 3 odd days.
? Last day of 2^nd century is Wednesday.
300 years contain ( 5 x 3) = 15 = 1 odd day.
? Last day of 3^rd century is Monday
400 years contain 0 odd day.
? Last day of 4^rd century is Sunday.
This cycle is repeated that means last day of century is Friday, Wednesday, Monday, or Sunday and it repeats the cycle.

Here we have two cases :
Case (i) : If a month start with Thursday the month will have 5 Saturdays and 4 Sundays i.e total 9 weekends. In this case 31^st December will be Sunday hence 1^st January will be either Sunday (For non leap year) or Saturday ( For leap year) then January will have 9 or 10 weekends.
Case (ii) : If a month start with Sunday the month will have 4 Saturdays and 5 Sundays i.e total 9 weekends.
In this case 31^st December will be Wednesday hence 1^st January will be either Wednesday (For non leap year) or Tuesday ( For leap year) then January will have 8 or 9 weekends.
Hence number of weekends in January will be 8 or 9 or 10.