RSA public key and private key calculation: multiple answer Given N = 91
public key (11, 91)
Plaintext (88)

Ciphertext = 88 ^11 mod 91 = 30

to decrypt, 
30 ^ 59 mod 91 = 88 (plaintext)
30 ^ 11 mod 91 = 88 (plaintext)
30 ^ 47 mod 91 = 88 (plaintext)


How can there be multiple suitable instance of private key such as 59/11/47?

Lastly, must public and private key always be a prime number?
[M]ust public and private key always be a prime number?


No, they don't have to be. Usually the public key is chosen as an odd prime though because it must hold that $\gcd(e,\lambda(n))=1$, which simplifies to "$e$ must not divide $\lambda(n)$" if $e$ is prime.


  How can there be multiple suitable instance of private key such as
  59/11/47?


Recall that all we need for correctness for RSA is that for each possible $m$ it holds that $m^x\bmod n=m$. It turns out that the smallest possible $x$ here isn't the euler-phi function $\varphi(n)=(p-1)(q-1)$ (in the two-prime case) but rather the carmichael function $\lambda(n)=\operatorname{lcm}(p-1,q-1)$ (in the two-prime case), because it is defined to be the smallest such $x$.

Now what does this mean? It means that if $ed\equiv 1\pmod{\lambda(n)}$ then $e,d$ is a valid pair of RSA exponents. As you can individually reduce each factor in modular arithmetic this also means that for a fixed choice of $e,d$ for any integers $k,l$ the following is also a pair of valid exponents: $e+l\lambda(n),d+k\lambda(n)$ because $(e+l\lambda(n))(d+k\lambda(n))=ed+ek\underbrace{\lambda(n)}_{0}+dl\underbrace{\lambda(n)}_{0}+\underbrace{\lambda(n)}_{0}\lambda(n)\bmod\lambda(n)=ed$. What you have experienced was the choice $e=11,d=11$ and $k=0,3,4$.