Rom

are you asking how many bingo cards have a top left to bottom right diagonal whose values are strictly increasing?

very clever dr. Jones!

This question doesn't make a ton of sense.

a is clearly less than 1 and thus a divided by any integer m will be 0m + a,

i.e. a itself will be the remainder.  Are you sure a isn't supposed to just be 3²ⁿ + 4?

f(1) is the sum of the coefficients of any polynomial f(x)

\(|z-3i|\leq 1\\ \text{describes a disk centered at }(0,3i) \text{ of radius }1\\ 0\leq \arg(z-3i)\leq \dfrac \pi 2\\ \text{says to only consider the first quadrant of that circle}\\ \text{The full circle will have area }\pi (1)^2 = \pi\\ \text{The first quadrant of this circle will have area }\dfrac \pi 4\)

There's no lack of videos on youtube on how to calculate the divergence and curl of a vector field.

Conservative vector fields have 0 curl.

CPhil is probably going to give you the proper answer so I'll chime in with a method that's probably

above where you're currently at but is one of the more clever tricks in the toolbox

.\(\cos(5\theta)= Re(e^{i5\theta}) = \\ Re((e^{i\theta})^5) = Re((\cos(\theta)+i\sin(\theta))^5)\\ \text{Now apply the binomial expansion}\\ Re\left(\sum \limits_{k=0}^5\dbinom{5}{k}\cos^k(\theta)(i\sin(\theta))^{5-k}\right)\)

\(\text{We can see the only terms that are real will be }k=1,3,5\\ \text{and we end up with}\\ \cos(5\theta) = \cos^5(\theta)-10\cos^3(\theta)\sin^2(\theta)+5\cos(\theta)\sin^4(\theta)\)

\(\text{Oh, I see you need it entirely in }\cos(\theta)\text{. Well we just do this}\\ \cos(5\theta) = \cos^5(\theta)-10\cos^3(\theta)(1-\cos^2(\theta))+ 5\cos(\theta)(1-\cos^2(\theta))^2\)

You can do the final simplification.

A is trivial.  This sort of problem is one of the first things you learn in discrete math class.  Think permutations.

B is less so. 

Each bracelet can be tagged with a base 4 digit indicating to which person it's assigned.

There are 4 bracelets so the different assignments can be represented by a 4 digit base 4 number.

There are 4⁴= 256 of these numbers and hence 256 different assignments.

The integration region is shown in purple.  The bottom curve is f(x) = x^1/3

The simple way to evaluate this integral is to switch order of integration and do.

\(\displaystyle \int_0^1\int_0^{y^3}~e^{y^4}~dx~dy\)

This integral should be pretty simple.  Give it a try.