Question 848062
Let X be a binomial(p=.10, n = 5)

P[X<=3]:

Two options. I suggest using the one that involves less calculations.

P[X<=3]: P[X=0] + P[X=1] + P[X=2] + P[X=3]
or
Using the complement of P[X<=3]: 1-P[X>=4] = 1-(P[X=4]+P[X=5])

I will use the 2nd.

P[X=4] = (5 choose 4) * (.10)^4 * (.90)^1 = 0.00045
P[X=5] = (5 choose 5) * (.10)^5 * (.90)^0 = 0.00001

1 - (0.00045 + 0.00001) = {{{highlight(.9995)}}}

Question 2:

P[3<=X<=5] =  P[X=3] + P[X=4] + P[X=5]

or  1-(P[X=0] + P[X=1] + P[X=2])

The first offers less calculations, and we've already calculated 2/3 of it in the first one. All that is required is to calculate 

P[X=3] = (5 choose 3) * (.10)^3 *(.90)^2 = .0081

P[X=3 or 4 or 5] = .0081 + .00045 + .00001 = {{{highlight(.0086)}}}

Just to make sure, I'll calculate using the complement as well.

P[X=0] = (5 choose 0)*(.10)^0*(.90)^5 = .5905

P[X=1] = 5 * .10^1  * .9^4 = .3281

P[X=2] = 10 * .10^2 * .9^3 = .0729

1 - .5905 - .3281 - .0729 =  {{{highlight(.0085)}}} <-close enough due to some rounding