**Solution:**

**a)**

**n^3+7(n^1..5)-3 and n^(log 8 -10(n^0.5)+7)**

we already know how Big- O, Big-Omega, Theta works but here is an overview to go through once again,

By looking at functions we can observe that

n^3+7(n^1..5)-3 <= c1* ( n^(log 8 -10(n^0.5)+7))

and Please note that log_{2} 8= log 2^3= 3 log 2= 3*1= 3, so n^log 8= n^3

n^3+7(n^1..5)-3 >= c2* (n^(log 8 -10(n^0.5)+7))

So we can say that

n^3+7(n^1..5)-3= (n^(log 8 -10(n^0.5)+7) )

Now,

**n log n and n^1.1**

Now we will compare by putting very large value of n and see the relation between these two

if n=2^256, then n log n= 2.9642775e+79 and

n^1.1= 5.8890708e+84

this means that n^1.1 is greater than n log n

Hence n log n= O(n^1.1).

**5^n and 5.1^n**

Only by looking at the function we can say that

5^n= o(5.1^n)

**4^n and n!**

We know that n! is almost equivalent to n^n

So here we can say that 4^n= o(n!)

**b)**

When we write generalized form of asymptotic notation then we only consider very large value of n, so the function which is taken at n>50 is going to be considered here,

f(n)= n log n + 100n

n log n + 100n <= c * (n log n )

for c= 2.

So, **f(n)= O(n log n)**