No, the vector $(\hat{i}+\hat{j})$ is not a unit vector because its magnitude is,
\[|\hat{i}+\hat{j}|=\sqrt{|i|^2+|j|^2+2|i||j|\cos90°}\] \[=\sqrt{1^2+1^2+0}=\sqrt{2}\]
Since its magnitude is equal to $\sqrt{2}$ not $1$, it is not a unit vector.
No, the vector $(\hat{i}+\hat{j})$ is not a unit vector because its magnitude is,
\[|\hat{i}+\hat{j}|=\sqrt{|i|^2+|j|^2+2|i||j|\cos90°}\] \[=\sqrt{1^2+1^2+0}=\sqrt{2}\]
Since its magnitude is equal to $\sqrt{2}$ not $1$, it is not a unit vector.