A few (very detailed) hints. A standard result for inner products is the following:
Theorem: Let $\mathcal{B}=\left\{\mathbf{x_1},\ \cdots,\ \mathbf{x_n}\right\}$ be an orthonormal set. Then if $\mathbf{y}$ is a linear combination of the vectors of $\mathcal{B}$ we have $\mathbf{y} = \langle\mathbf{y},\mathbf{x_1}\rangle\mathbf{x_1} + \cdots + \langle\mathbf{y},\mathbf{x_n}\rangle\mathbf{x_n}$ Where $\langle\cdot,\cdot\rangle$ is the associated inner product.
Proof: Write $\mathbf{y}$ as a linear combination of the $\mathbf{x_i}$s. We have $\mathbf{y} = c_1\mathbf{x_1} + \cdots + c_n\mathbf{x_n}$ for undetermined coefficients $c_i$. Taking the inner product of both sides with each $\mathbf{x_i}$ and exploiting orthonormality, we have $\langle\mathbf{y},\mathbf{x_i}\rangle = \langle\mathbf{x_i},\mathbf{x_i}\rangle c_i = c_i$ The result follows. $\square$
Can you see how the method of this proof directly solves your problem? Even more simply, you can just apply the results of this theorem to question b.
For c, consider how you would calculate $\|\mathbf{y}\|^2$. You would have $\|\mathbf{y}\|^2 = \langle\mathbf{y},\mathbf{y}\rangle = \langle c_1\mathbf{x_1}+c_2\mathbf{x_2},\ c_1\mathbf{x_1}+c_2\mathbf{x_2}\rangle$ Using properties of inner products, how can you simplify this? Note that the method you would take to solve the above problem also generalizes to an arbitrary number of orthonormal vectors. It is the Pythagorean theorem in $n$-dimensions.