How many integer solutions are there to the equation $x_1 + x_2 + x_3 + x_4 = 12$ with restrictions on $x_1,x_2,x_3,x_4$? How many integer solutions are there to the equation $x_1 + x_2 + x_3 + x_4 = 12$ with $x_i > 0$ for each $i \in \{1, 2, 3, 4\}$?  How many solutions with $x_1 > 1$, $x_2 > 1$, $x_3 > 3$, $x_4 \geq 0$?
So for the first part I did this:
We set $x_i = y_i + 1$ for $i = 1, 2, 3$ and obtain $y_1 + y_2 + y_3 = 12 − 3 = 9$.
So, there is one-to-one correspondence between the positive integer solutions of $x_1 + x_2 + x_3 = 12$ and the non-negative integer solutions of $y_1 + y_2 + y_3 = 9$.
Hence, both equation have the same number of solutions. Since $y_1 + y_2 + y_3 = 9$ has ${9+3-1 \choose 3-1} = {11 \choose 2} = 55$ integer solutions.
Is this correct?
For the second part I do not understand how to figure out how many solutions with $x_1 > 1$, $x_2 > 1$, $x_3 > 3$, $x_4 \geq 0$? 

I am not sure why you have four variables in the statement of your question and three variables in your answer.  I will assume you meant to work with four variables.  

How many integer solutions are there to the equation $x_1 + x_2 + x_3 + x_4 = 12$ with $x_i > 0$ for each $i \in \{1, 2, 3, 4\}$?

We wish to solve the equation 
$$x_1 + x_2 + x_3 + x_4 = 12 \tag{1}$$
in the positive integers.  
Method 1:  We reduce the problem to one in the non-negative integers. Let $y_k = x_k - 1$ for $1 \leq k \leq 4$.  Then each $y_k$ is a non-negative integer.  Substituting $y_k + 1$ for $x_k$, $1 \leq k \leq 4$, in equation 1 yields
\begin{align*}
y_1 + 1 + y_2 + 1 + y_3 + 1 + y_4 + 1 & = 12\\
y_1 + y_2 + y_3 + y_4 & = 8 \tag{2}
\end{align*}
Equation 2 is an equation in the non-negative integers.  A particular solution corresponds to placing three addition signs in a row of eight ones.  For instance, 
$$1 1 1 1 1 + 1 + 1 1 1$$
corresponds to the solution $y_1 = 5$, $y_2 = 1$, and $y_3 = 3$, while 
$$1 1 + + 1 1 1 1 1 1$$
corresponds to the solution $y_1 = 2$, $y_2 = 0$, and $y_3 = 6$.  Thus, the number of solutions of equation 2 is the number of ways three addition signs can be inserted into a row of eight ones, which is 
$$\binom{8 + 3}{3} = \binom{11}{3}$$
since we must choose which three of the eleven symbols (eight ones and three addition signs) will be addition signs.
Method 2:  A particular solution of equation 1 in the positive integers corresponds to inserting three addition signs in the eleven spaces between successive ones in a row of $12$ ones. 
$$1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1 \square 1$$
For instance, 
$$1 1 1 + 1 1 1 1 + 1 1 1 1 1$$
corresponds to the solution $x_1 = 3$, $x_2 = 4$, and $x_3 = 5$.  Thus, the number of solutions of equation 1 in the positive integers is the number of ways three addition signs can be inserted into the eleven gaps between successive ones in a row of $12$ ones, which is 
$$\binom{11}{3}$$

How many integer solutions are there to the equation $x_1 + x_2 + x_3 + x_4 = 12$ with $x_1 > 1$, $x_2 > 1$, $x_3 > 3$, $x_4 \geq 0$?

We reduce the problem to one in the non-negative integers.  Since $x_1$ is an integer, $x_1 > 1 \implies x_1 \geq 2$.  Similarly, since $x_2$ and $x_3$ are integers, $x_2 > 1 \implies x_2 \geq 2$ and $x_3 > 3 \implies x_3 \geq 4$.  Let 
\begin{align*}
y_1 & = x_1 - 2\\
y_2 & = x_2 - 2\\
y_3 & = x_3 - 4\\
y_4 & = x_4
\end{align*}
Then each $y_k$, $1 \leq k \leq 4$, is a non-negative integer.  Substituting $y_1 + 2$ for $x_1$, $y_2 + 2$ for $x_2$, $y_3 + 4$ for $x_3$, and $y_4$ for $x_4$ in equation 1 yields
\begin{align*}
y_1 + 2 + y_2 + 2 + y_3 + 4 + y_4 & = 12\\
y_1 + y_2 + y_3 + y_4 & = 4 \tag{3}
\end{align*}
Equation 3 is an equation in the non-negative integers with 
$$\binom{4 + 3}{3} = \binom{7}{3}$$
solutions.
