Are all finite fields isomorphic to $\mathbb{F}_p$? I've recently started taking some algebra courses and I was wondering whether or not every finite field is isomorphic to $\mathbb{F}_p$, where $p$ is prime.
 A: Not quite. 

There is a finite field with $q$ elements, iff $q=p^k$ for some prime $p$. This field is unique up to isomorphy. 

So finite fields with a prime number of elements are indeed isomorphic to $F_p$. 
But there are also fields with a prime power number of elements, usually also denoted by $F_q$ where $q=p^k$, $k>1$. They can be constructed as following: 
$$F_q\simeq F_p[X]/(f(X))$$
where $f\in F_p[X]$ is an irreducible polynomial of degree $k$. This construction does not depend on the choice of  $f$, as different polynomials (of the same degree) lead to isomorphic fields.
A: No, every finite field is isomorphic to $\mathbb F_q$ where $q = p^n$ for some prime $p$ and non-negative integer $n$.
A: No, not all finite fields are isomorphic to a field of prime cardinality. What is however true is that the cardinality of a finite field is always a power of a prime. And, up to isomorphism there is a unique field of cardinality $q$ for each prime power $q$. 
A: No. But every finite field is isomorphic to $$F_{p^n}$$ for a prime $p$ and a positive integer $n$. In particular, the finite field with $p^n$ elements is the splitting field of the polynomial
$$x^{p^n} - x$$
over $F_p$. There's a full Wikipedia article here about these.
A: You're close, every finite field is isomorphic to $\mathbb{F}_{p^n}$, which is a quotient of $\mathbb{F}_p[x]$ by an irreducible polynomial of degree $n$.  It's often difficult to see that things like this can exist without a concrete example.  The easiest is $\mathbb{F}_4$:
$$\begin{array}{c|cccc}
+       &   0       &   1       &   \alpha  &   1+\alpha\\ \hline
0       &   0       &   1       &   \alpha  &   1+\alpha\\
1       &   1       &   0       &   1+\alpha&   \alpha  \\
\alpha  &   \alpha  &   1+\alpha&   0       &   1       \\
1+\alpha&   1+\alpha&   \alpha  &   1       &   0       \\
\end{array}$$
$$\begin{array}{c|cccc}
\times  &   \color{grey}{0}       &   1       &   \alpha  &   1+\alpha\\ \hline
\color{grey}{0}       &   \color{grey}{0}       &   \color{grey}{0}       &   \color{grey}{0}       &   \color{grey}{0}       \\
1       &   \color{grey}{0}       &   1       &   \alpha  &   1+\alpha\\
\alpha  &   \color{grey}{0}       &   \alpha  &   1+\alpha&   1       \\
1+\alpha&   \color{grey}{0}       &   1+\alpha&   1       &   \alpha  \\
\end{array}$$
You can see that $\mathbb{F}_4$ is a group under addition and $\mathbb{F}_4^\times=\mathbb{F}_4\setminus \{0\}$ is a group under multiplication.
In this case, I made $\mathbb{F}_4=\mathbb{F}_{2^2}$ out of the polynomial $x^2+x+1$, which is irreducible over $\mathbb{F}_2$.  I get the multiplication table by taking all polynomials of degree $1$ or less (i.e. below the degree of the irreducible polynomial minus one), then multiplying them together and reducing them mod $x^2+x+1$. (For calculation, it doesn't matter whether you use $x$ or $\alpha$.)
A: No. As the wikipage linked to by Jack explains there is a unique (up to isomorphism) finite field $\mathbb{F}_q$ with $q$ elements whenever $q=p^n$ is a power of a prime $p$ ($n$ a positive integer).
The simplest example is $\mathbb{F}_4$. It has elements $\{0,1,\alpha,1+\alpha\}$, where $\alpha$ is a mysterious constant that satisfies the equation
$$\alpha^2=\alpha+1.\qquad(*)$$
For all $x\in\mathbb{F}_4$ we have $x+x=0$. This explains everything there is to know about the addition in $\mathbb{F}_4$, so for example
$$
\alpha+(1+\alpha)=1+(\alpha+\alpha)=1+0=1.
$$
That mysterious equation $(*)$ explains everything you need to know about multiplication in $\mathbb{F}_4$ (together with ring axioms). So for example
$$
\alpha(1+\alpha)=\alpha+\alpha^2=\alpha+(1+\alpha)=1.
$$
This means that $\alpha$ and $1+\alpha$ are (multiplicative) inverses to each other. From this it follows easily that $\mathbb{F}_4$ is, indeed, a field.
An exercise for you: Show that $x^3=1$ for all non-zero elements $x\in\mathbb{F}_4$.
