How do I find the probability of committing a Type II error? I'm not sure If I understand what (b) is asking. Does it mean that the alternative hypothesis will be p<0.3, p<0.4, and p<0.5? If it is, then I have to find the probability of committing a type 2 error for each alternatives? If I am right then the setup that I have for p=0.3 in part (b) is correct.

Click here to see the question and my worked out problem
 A: You are on the right track, but you just need to go through and complete the calculation.  Note that the sum $$\Pr[X > 3 \mid p = 0.3] = \sum_{x=4}^{10} \binom{10}{x} (0.3)^x (0.7)^{10-x}$$ contains a total of $10-4+1 = 7$ terms, whereas you could save work by exploiting the complementary probability:  $$\Pr[X > 3 \mid p = 0.3] = 1 - \Pr[X \le 3 \mid p = 0.3] = 1 - \sum_{x=0}^3 \binom{10}{x} (0.3)^x (0.7)^{10-x},$$ and this latter sum has only $4$ terms, and they are easy to compute:  $$\sum_{x=0}^3 \binom{10}{x} (0.3)^x (0.7)^{10-x} = (0.7)^{10} + 10(0.3)(0.7)^9 + 45(0.3)^2(0.7)^8 + 120(0.3)^3(0.7)^7.$$
A: In R statistical software, dbinom is the binomial PDF with
parameters specified by appropriate arguments. Your computations
of Type I and II error probabilities can be coded as shown below.
 dbinom(0:3,10,.6)
 ## 0.0001048576 0.0015728640 0.0106168320 0.0424673280
 sum(dbinom(0:3,10,.6))
 ## 0.05476188

 sum(dbinom(4:10,10,.3))
 ## 0.3503893
 sum(dbinom(4:10,10,.4))
 ## 0.6177194
 sum(dbinom(4:10,10,.5))
 ## 0.828125

When computing these values via the binomial PDF with a
calculator, it is easiest to sum the fewest possible terms,
subtracting from 1 when appropriate, as in the helpful Answer by @heropup. 
The answers above from R will give  you a way to check
the values you get by  hand.
The probability of Type II error increases as the alternative value of $p = P(\text{Late})$
gets closer to the hypothetical value of $p$.
The 'power' of the test against alternative $p_a,$
sometimes denoted as $\pi(p_a),$ is the probability of rejecting the null hypothesis when it is false: $\pi(p_a) = 1 - \beta(p_a).$
That is, the power is $1 - P(\text{Type II Error}),$
for a particular alternative value of $p.$ So $$\pi(.3) = 1 - 0.3504 = 0.6496 > \pi(.5) = 1 - 0.8281 = 0.1719.$$
The plots below provide an overview of Type I and II error
probabilities for $p = .6$ and $p = .3,$ respectively. The
critical value is represented by a dotted vertical purple line,
with the rejection region to the left of it.

