In how many ways can a student score exactly $100$ points on four $50$ point exams? greater than $100$?

To score exactly 100, first suppose there are no constraints other than having nonnegative numbers. Then the number of solutions to $a+b+c+d = 100$ is simply $\binom{103}{3}$ by a standard stars and bars argument. Now we need only remove solutions where one of the values is greater than $50.$ Suppose $ a > 50,$ then we ask for solutions to $(a-51) + b + c + d = 49,$ where $a,b,c,d$ are nonnegative, which has $\binom{52}{3}$ solutions. Since we could have had any of the four be the one larger than $50,$ there are $4\cdot \binom{52}{3}$ such "bad" solutions. Hence the required total is $N_{100} = \binom{103}{3} - 4\binom{52}{3}.$

For the second part, any solution to $a + b + c +d > 100$ corresponds with $(50-a) + (50-b) + (50-c) + (50-d) < 100.$ Hence the number of such solutions is just $(51^4 - N_{100})/2,$ where $N_{100}$ is the answer to the first part

Here is one brute-force approach which is less elegant than the other answer but perhaps will showcase a different way of arriving at the same number.

Let's split up our exams into two groups. Exam one and two are in group $A$ and exam three and four are in group $B$. Let $50 \leq n \leq 100$. We will want to consider the number of ways that $n$ points are obtained in group $A$ and the remaining $(100 - n)$ in group $B$.

First, suppose $n = 50$. The number of ways for group $A$ to score $50$ is $51$, since there is one possibility for each outcome of exam one, ranging from $0$ to $50$. There are a similar number of possibilities for group $B$. Hence $(51)^2$ total possibilities exist in this case.

Otherwise, one group scores above $50$ and the other below $50$. Suppose that it is group $A$ that scores higher -- we can justify this assumption by multiplying by $2$ at the end, as there are identical cases when group $B$ scores higher. For $n>50$ there are $(101 - n)$ ways for group $A$ to score $n$ points. For example if $n = 100$ then there is one way, and if $n = 60$ there are $41$ possibilities, as exam one can receive between $10$ and $50$ points. Next, with the remaining $(100-n)$ points for group B there are $(100-n)+1 = 101-n$ possibilities. This is because, for example, if $n=60$ then $100-n = 40$ and there are $41$ possibilities, as exam $3$ can score between $0$ and $40$ points. Thus the total number of possibilities are $(101 -n)^2$. This formula is nice and symmetric, and suggests that a more beautiful analysis than the one here is possible. Anyhow, we get the sum total number of possibilities as

$$(51)^2 + 2\sum_{n=51}^{100} (101 - n)^2 = (51)^2 + 2\sum_{n=1}^{50} n^2$$

Using a simple formula for the sum of squares, we get $(51)^2 + 2(42925) = 88451$.

So far so good. How do we compute the total number of ways of score above a hundred? By symmetry, this is equal to the total number of ways of getting less than $100$, simply by replacing every right answer by a wrong answer. So the total number of possible scores $(51)^4$ is the sum of $88451$ and twice the score in question. We compute

$$ \frac{51^4 - 88451}{2} = 3338375$$.

The scores he can get are the coefficients of the different powers of $x$ in the expansion \begin{align*} (1+x+\cdots + x^{50})^4 &= \left(\frac{1-x^{51}}{1-x}\right)^4 \\ &= (1-4x^{51} + 6X^{102} - 4X^{153} + X^{204})\left(1+ 4x + \frac{4\cdot 5}{1\cdot 2}x^2 + \frac{4\cdot 5 \cdot 6}{3!}x^3 + \cdots +\binom{n+3}{3}x^n+\cdots \right) \end{align*} Hence the number of ways in which he can get exactly 100 is $$ \binom{103}{3} - 4 \binom{52}{3} = 88451$$ Second part has already been solved in the other solutions posted.