$\hat{i}$ and $\hat{j}$ are unit vectors along x- and y- axis respectively.

A) What is the magnitude and direction of the vectors $\hat{i}+\hat{j}$ and $\hat{i}-\hat{j}$ ?

B) What are the components of a vector $\overrightarrow{A}=2\hat{i}+3\hat{j}$ along the directions of $\hat{i}+\hat{j}$, and $\hat{i}-\hat{j}$ ?

A) Part 1 : - Recall how to find magnitude and direction of a vector.
Let us assume $\overrightarrow{C}=\hat{i}+\hat{j}=C_x\hat{i}+C_y\hat{j}$
$|\overrightarrow{C}|=\sqrt{(1{)}^2+(1{)}^2}=\sqrt{2}\text{unit}$
Direction from x-axis:
$\tan \theta =\frac{C_y}{C_x}=1$
$\theta ={45}^{\circ }$

Similarly, for $\hat{i}-\hat{j}$
Assume $\overrightarrow{B}=B_x\hat{i}+B_y\hat{j}=\hat{i}-\hat{j}$
$|\overrightarrow{B}|=\sqrt{B_x^2+B_y^2}=\sqrt{(1{)}^2+(-1{)}^2}=\sqrt{2}\text{unit}$
Direction of B from x-axis:
$\tan \varphi =\frac{B_y}{B_x}=-1$
$\varphi ={45}^{\circ }$ in clockwise sense.

B) Part 2 : - Component of one vector in the direction of another vector.
If $\overrightarrow{A}\&\overrightarrow{B}$ are two vectors and $\theta$ is the angle between them, then
$\overrightarrow{A}.\overrightarrow{B}=|\overrightarrow{A}||\overrightarrow{B}|\cos \theta$
Component of $\overrightarrow{A}$ along $\overrightarrow{B}$ is, $|\overrightarrow{A}|\cos \theta =\frac{\overrightarrow{A}.\overrightarrow{B}}{|\overrightarrow{B}|}$
So, component of $\overrightarrow{A}=2\hat{i}+3\hat{j}$ along the direction of $\overrightarrow{C}=\hat{i}+\hat{j}$,
$A\cos \theta =\frac{\overrightarrow{A}.\overrightarrow{C}}{|\overrightarrow{C}|}=\frac{(2\hat{i}+3\hat{j}).(\hat{i}+\hat{j})}{\sqrt{(1{)}^2+(1{)}^2}}$
$A\cos \theta =\frac{2+3}{\sqrt{2}}=\frac{5}{\sqrt{2}}\text{unit}$.
Similarly, component of $\overrightarrow{A}=2\hat{i}+3\hat{j}$ along the direction of $\overrightarrow{B}=\hat{i}-\hat{j}$,
$A\cos \theta =\frac{\overrightarrow{A}.\overrightarrow{B}}{|\overrightarrow{B}|}=\frac{(2\hat{i}+3\hat{j}).(\hat{i}-\hat{j})}{\sqrt{(1{)}^2+(-1{)}^2}}$
$=\frac{2-3}{\sqrt{2}}=-\frac{1}{\sqrt{2}}\text{unit}$.