The relation defining your space is
$$
X \in S \quad \Leftrightarrow \quad
\langle X, (6, -2, 4, -10) \rangle = 0
$$
where $\langle \cdot, \cdot \rangle$ is the dot product. So one very obvious guess of a vector that is orthogonal to all $X$ in $S$ is $(6, -2, 4, -10)$. The orthogonal complement of $S$ is, therefore, the space generated by $u = (6, -2, 4, -10)$. (By dimension counting, you know that $1$ generator is enough.) The projection operation is
$$
P(X) = X - \frac{\langle X, u\rangle}{\langle u, u\rangle}u
= X - \frac{uu^T}{u^Tu}X = \left(I - \frac{uu^T}{u^Tu}\right)X.
$$
Recall that a linear transformation from $A^m\to B^n$, where $m,n$ are dimensions of $A,B$ respectively, has a matrix of order $n\times m$. Thus, $T:\Bbb R^4\to\Bbb R^4$ is a linear map having a $4\times4$ square matrix, $[T]$.
For showing that $T$ is not invertible, you can show that there is an element in $\ker(T)$ other than $(0,0,0,0)$, that is, $\dim(\ker(T))>0$. For example, $\mathbf u=(7,1,5,0)$ has $\bf0$ projection on $V$. Alternatively, you can solve $(b)$ first and show that the matrix of $T$ isn't invertible, that is, $\text{rank}([T])<4$.
For part $(b)$, let $\mathbf v=(x,y,z,w)\in\Bbb R^4$. The basis $B=\{\mathbf a,\mathbf c\}$ of $V$ consists of orthogonal linearly independent vectors, so the orthogonal projection of $\mathbf v=(x,y,z,w)$ on $V$ is the vector sum $$\displaystyle\frac{\langle\mathbf v,\mathbf a\rangle}{\langle\mathbf a,\mathbf a\rangle}\cdot\mathbf a+\frac{\langle\mathbf v,\mathbf c\rangle}{\langle\mathbf c,\mathbf c\rangle}\cdot\mathbf c$$
Note that you are only able to do this because the basis vectors are orthogonal.
This gives the orthogonal projection of $\mathbf v=(x,y,z,w)$ as
$\displaystyle=\Big[\frac{x-2y-z+3w}{15}\Big]\cdot(1,-2,-1,3)+\Big[\frac{2x+y-3z-w}{15}\Big]\cdot(2,1,-3,-1)\\\displaystyle=\Big(\frac{5x-7z+w}{15},\frac{5y-z-7w}{15},\frac{-7x-y+10z}{15},\frac{x-7y+10w}{15}\Big)$
Thus, the matrix of $T,[T]=\begin{bmatrix}1/3&0&-7/15&1/15\\0&1/3&-1/15&-7/15\\-7/15&-1/15&2/3&0\\1/15&-7/15&0&2/3\end{bmatrix}$.
Edit. As hinted by you in a comment, we can show that $T$ is not invertible by noting that the range $R(T)$ is two-dimensional, so nullity of $T>0$.
Best Answer
Fact: $P$ is a projection matrix iff $P^2 = P$.
So, we need to show that $P^2 = P \implies (I-P)^2 = I-P$.
Do you see how to do this?
EDIT: As mentioned by Vedran Šego in the comments below, the above only shows that $P$ is a projection matrix, not necessarily an orthogonal projection matrix. To show that $P$ is an orthogonal projection matrix, we also need to show that $P$ is symmetric $\implies$ $I-P$ is symmetric.