If the probability of success is "p" and the probabibility of failure is "q", then the probability of at least "k" successes on n trials is C(n, k) *
p^k * q^(n - k)

Using the above we find n = 100, p = q = 1/2 = 0.5 and,

A. H = 50,
P = C(100, 50) * (1/2)^50*(1/2)^(100-50) = C(100, 50)/2^100

B. T >= 60, P = C(100, 60) * (1/2)^40*(1/2)^60 + C(100, 61) * (1/2)^39 * (1/2)^ 61 + ...
P = [ C(100, 60) + C(100, 61) + C(100, 62) + .... + C(100, 100) ]/2^100

C. 51 <= H <= 55,
P = [ C(100, 51) + ... + C(100, 55) ] / 2^100

D. H >= 48 and T >=48
P = [ C(100, 48) + C(100, 49) + C(100, 50) + C(100, 51) + C(100, 52) ] / 2^100

E. H <=5 or H >=95
P = [ C(100, 0) + C(100, 1) + ... + C(100, 5) + C(100, 95) + ... + C(100, 100) ] / 2^100

Now using the properties of C(n, k) = n!/[k!(n - k)!], the above may be somewhat simplified to,

A. P = C(100, 50)/2^100

B. P = [ C(100, 60) + C(100, 61) + C(100, 62) + .... + C(100, 100) ]/2^100
= 1 - [ C(100, 0) + ... C(100, 49) + C(100, 50) + C(100, 51) + ... + C(100, 59) ] / 2^100

C. P = [ C(100, 51) + ... + C(100, 55) ] / 2^100

D. P = [ C(100, 48) + C(100, 49) + C(100, 50) + C(100, 51) + C(100, 52) ] / 2^100

E. P = 2 [ C(100, 0) + C(100, 1) + ... C(100, 5) ] / 2^100

Which one is the greatest? Well to be precise use a calculator. Or...
Observe that C > A, D > A, E can easily be calculated to be small, B although large is far away from the middle value C(100, 50).

Elementarily, I know that the combinatorial function takes its greatest value at the midpoint.
E.g. C(2, 0) = 1, C(2, 1) = 2, C(2, 2) = 1
C(7, 0) = 1, C(7, 1) = 7, C(7, 2) = 21, C(7, 3) = 35, C(7, 4) = 35, ....

So I would bet on D. as it takes the largest number of values from the middle of the range for C(100, k).