Now that you know how to calculate DFT and IDFT and what they do, time for some practice in calculating their values.
Before doing calculations let's derive a formula for complex sinusoid with negative exponential that is used every time you compute DFT.
$$ e^{-j*2\pi*k/N*n} = cos(-2\pi*k/N*n) + j*sin(-2\pi*k/N*n) = \\ cos(2\pi*k/N*n) - j*sin(2\pi*k/N*n) $$
For making transition above I have used symmetry properties of sine and cosine that are
$$cos(-x) = cos(x)$$
$$sin(-x) = -sin(x)$$
Let's say we have a signal $x = (1,2)$. So number of samples in the signal is $N=2$. This is our first time doing calculations so let's take everything in baby steps.
For remainder
$$DFT(x) = y = (y_0, y_1, y_2, ..., y_{N-1})$$
$$y_k = \sum_{n=0}^{N-1}x_n*e^{-j*2\pi*k/N*n}$$
Because $k=0,1,...,N-1$, there are only two coefficents to compute: $y_0$ and $y_1$.
$$y_0 = \sum_{n=0}^{2-1}x_n*e^{-j*2\pi*0/2*n} = \sum_{n=0}^{1}x_n*e^{0} = \sum_{n=0}^{1}x_n*1 = x_0 + x_1 = 1 + 2 = 3$$
$$y_1 = \sum_{n=0}^{2-1}x_n*e^{-j*2\pi*1/2*n} = \sum_{n=0}^{1}x_n*e^{-j*2\pi*1/2*n} = x_0 e^{-j*2\pi*1/2*0} + x_1 e^{-j*2\pi*1/2*1} = \\ x_0 e^{0} + x_1 e^{-j*2\pi*1/2} = x_0 * 1 + x_1 (cos(2\pi * 1/2) - jsin(2\pi * 1/2)) = \\ x_0 + x_1 (-1 - j * 0) = x_0 - x_1 = 1 - 2 = -1$$
So $DFT(x) = DFT((1,2)) = y = (3, -1)$.
And now the other way.
$$IDFT(y) = x = (x_0, x_1, x_2, ..., x_{N-1})$$
$$x_n = \frac{1}{N} \sum_{k=0}^{N-1}y_k * e^{j*2\pi*k/N*n}$$
$n = 0,1,...,N-1$, so there are only two samples to compute: $x_0$ and $x_1$.
$$x_0 = \frac{1}{2} \sum_{k=0}^{2-1}y_k * e^{j*2\pi*k/2*0} = \frac{1}{2} \sum_{k=0}^{1}y_k * e^{0} = \frac{1}{2} \sum_{k=0}^{1}y_k * 1 = \frac{1}{2} (y_0 + y_1) = \\ \frac{1}{2} (3 + (-1)) = \frac{1}{2} * 2 = 1 $$
$$x_1 = \frac{1}{2} \sum_{k=0}^{2-1}y_k * e^{j*2\pi*k/2*1} = \frac{1}{2} \sum_{k=0}^{1}y_k * e^{j*2\pi*k/2} = \frac{1}{2} (y_0 * e^{j*2\pi*0/2} + y_1 * e^{j*2\pi*1/2}) = \\ \frac{1}{2} (y_0 * e^{0} + y_1 * (cos(2\pi*1/2) + jsin(2\pi*1/2)) ) = \\ \frac{1}{2} (y_0 * 1 + y_1 * (-1 + j*0) ) = \frac{1}{2} (y_0 - y_1) = \\ \frac{1}{2} (3 - (-1)) = \frac{1}{2} * 4 = 2 $$
So $IDFT(y) = IDFT((3,-1)) = x = (1,2)$ which matches with original signal $x$.
Let our signal be $x=(6,3,5,1)$. There are 4 samples in the signal so $N=4$. This time I will show only key points during computation.
Forward DFT
$$y_k = \sum_{n=0}^{N-1}x_n*e^{-j*2\pi*k/N*n}$$
$$y_0 = \sum_{n=0}^{4-1}x_n*e^{-j*2\pi*0/4*n} = x_0 + x_1 + x_2 + x_3 = 15$$
$$y_1 = \sum_{n=0}^{4-1}x_n*e^{-j*2\pi*1/4*n} = x_0 * 1 + x_1 * (-j) + x_2 * (-1) + x_3 * j = 1 - 2j$$
$$y_2 = \sum_{n=0}^{4-1}x_n*e^{-j*2\pi*2/4*n} = x_0 * 1 + x_1 * (-1) + x_2 * 1 + x_3 * (-1) = 7$$
$$y_3 = \sum_{n=0}^{4-1}x_n*e^{-j*2\pi*3/4*n} = x_0 * 1 + x_1 * j + x_2 * (-1) + x_3 * (-j) = 1 + 2j$$
$DFT((6,3,5,1)) = (15, 1-2j, 7, 1+2j)$.
Let's have a closer look at how $y_1$ is computed to make sure you understand everything that is going on.
$$y_1 = \sum_{n=0}^{4-1}x_n*e^{-j*2\pi*1/4*n} = \sum_{n=0}^{3}x_n*e^{-j*2\pi*1/4*n} = \\ x_0*e^{-j*2\pi*1/4*0} + x_1*e^{-j*2\pi*1/4*1} + x_2*e^{-j*2\pi*1/4*2} + x_3*e^{-j*2\pi*1/4*3} = \\ x_0 (cos(2\pi*1/4*0) - jsin(2\pi*1/4*0)) + \\ x_1 (cos(2\pi*1/4*1) - jsin(2\pi*1/4*1)) + \\ x_2 (cos(2\pi*1/4*2) - jsin(2\pi*1/4*2)) + \\ x_3 (cos(2\pi*1/4*3) - jsin(2\pi*1/4*3)) = \\ x_0 (1-j*0) + x_1 (0-j*1) + x_2 (-1-j*0) + x_3 (0-j*(-1)) = \\ x_0 * 1 + x_1 * (-j) + x_2 * (-1) + x_3 * j = \\ 6 * 1 + 3 * (-j) + 5 * (-1) + 1 * j = 6 -3j -5 +j = 1 -2j $$
If you are doing calculations yourself and have $y_1=1+2j$ and $y_3=1-2j$ then you are forgetting minus sign in exponential or before sin().
And inverse DFT.
$$x_n = \frac{1}{N} \sum_{k=0}^{N-1}y_k * e^{j*2\pi*k/N*n}$$
$$x_0 = \frac{1}{4} \sum_{k=0}^{4-1}y_k * e^{j*2\pi*k/4*0} = \frac{1}{4} (y_0 + y_1 + y_2 + y_3) = 6$$
$$x_1 = \frac{1}{4} \sum_{k=0}^{4-1}y_k * e^{j*2\pi*k/4*1} = \frac{1}{4} (y_0 * 1 + y_1 * j + y_2 * (-1) + y_3 * (-j)) = 3$$
$$x_2 = \frac{1}{4} \sum_{k=0}^{4-1}y_k * e^{j*2\pi*k/4*2} = \frac{1}{4} (y_0 * 1 + y_1 * (-1) + y_2 * 1 + y_3 * (-1)) = 5$$
$$x_3 = \frac{1}{4} \sum_{k=0}^{4-1}y_k * e^{j*2\pi*k/4*3} = \frac{1}{4} (y_0 * 1 + y_1 * (-j) + y_2 * (-1) + y_3 * j) = 1$$
$IDFT((15, 1-2j, 7, 1+2j)) = (6,3,5,1)$
You cannot learn to do DFT just by reading about it just as your body won't get fit by reading about exercises. Here I have put some signals and their transforms so you can compute them yourself and check results.
$DFT((0,1)) = (1,-1)$
$DFT((2,5)) = (7,-3)$
$DFT((5,2)) = (7,3)$
$DFT((3+j,2-j)) = (5+0j, 1+2j)$
$DFT((3,2,4,1)) = (10+0j, -1-1j, 4+0j, -1+1j)$
$DFT((2,5,1,8)) = (16+0j, 1+3j, -10+0j, 1-3j)$