Find a single-sampling plan for which $p_1=0.02$, $\alpha=0.01$,$p_2=0.06$ and $\beta=0.10$ I am not sure where to start. My previous lecture gave very limited information on what to do here, so my current toolbox only consists of these formulas:
$$1-\alpha=\sum_{d=0}^{c} {n \choose d}p_1^d(1-p_1)^d$$
$$\beta=\sum_{d=0}^{c} {n \choose d}p_2^d(1-p_2)^d$$
where $d$ is defects and $c$ is our acceptance number. It seems quite painful to have to solve this for $n$ and $c$, so I presume there is a more efficient method. I am not sure what this is however. I looked this problem up and found the concept of a "binomial monograph", but this has not been taught in my class, thus I am not sure if it is applicable. 
Goal of the question: How to solve single-sampling plan problem using the given data and formulas, preferably within my current toolbox (knowledge that a college statistics major should have)
 A: It seems you have information from $n$ Bernoulli trials and wish
to test $H_0: p = 0.02$ against $H_a: p = 0.06$ at level $\alpha = 0.01.$
To begin, suppose $n=500.$ [This hypothesis testing situation is
called 'simple vs. simple' because one specific value is specified
for the null hypothesis and and another for the alternative.]
To start, let $n=500$ Bernoulli trials. Then, under the null hypothesis,
the distribution of the number $X$ of successes observed in 500 is
$X \sim \mathsf{Binom}(n=500, p=0.02.$ You want to choose a critical
value $c$ with $$P(\mathrm{Rej}|p=.02) = P(X \ge c) = 1 \le .01 = \alpha.$$ You should reject $H_0$ if you
see a relatively high number of successes.
in R statistical software [where dbinom is a binomial PDF, pbinom is a binomial CDF, and qbinom is a binomial quantile function (inverse CDF)], we have the computations below, which show that $c = 19.$ I don't know whether you are using technology such as R or whether you are expected to use normal approximations to binomial probabilities and use printed standard normal CDF tables to get probabilities.
qbinom(.99, 500, .02)  
[1] 18
sum(dbinom(18:500, 500,.02))
[1] 0.01339108
sum(dbinom(19:500, 500,.02))
[1] 0.00662783
1 - pbinom(18, 500, .02)
[1] 0.00662783

Then the next step is to see what value of $\beta$ arises from the above.
We want to know whether
$$\beta = P(\mathrm{No Rej}|H_a)  = P(X < c = 19|p = .06) < 0.1,$$
where now $X \sim \mathsf{Binom}(n =500, p-.06).$ So $n=500$ is certainly a large enough sample size because $\beta$ is much smaller than it needs to be.
pbinom(18, 500, .06)
[1] 0.01095657

I don't know the exact nature of your assignment. By trial and error in R you could find the minimal value of $n$ that would suffice to get values of $\alpha$ and $\beta$ that meet your specifications. Maybe you're
just supposed to get an $n$ of 'reasonable' size that does the job.
If you need the minimal $n,$ you might be able to find it by solving
equations that result from standardizing to use standard normal tables.
With this start, I will leave the rest to you.
Here is a plot of the null (blue) and alternative (maroon) binomial
 distributions. The black vertical line is at $c,$ as computed above. For $n = 500$ the overlap between them is somewhat less than it needs to be.

