= Solution
Use the standard nonnegative partition-normalized <B-splines> $N_j$; their sum is at most one, and is exactly one for the complete clamped <basis> on $[0,1]$. The printed question introduces $M_i$ only in its last <matrix> formula without defining it. The intended normalization is the <unit-integral normalization of a B-spline>:
$$
c_i=\int_0^1N_i(t)\,dt>0,\qquad M_i(t)=\frac{N_i(t)}{c_i},\qquad \int_0^1M_i(t)\,dt=1.
$$
For supports wholly inside the interval, $c_i=(t_{i+k}-t_i)/k$. The positivity and partition convention, sometimes called $L_\infty$ normalization, should not be confused with individually rescaling every <spline> to have maximum exactly one.
The characterization in part 1(a) gives the normal equations for the <orthogonal projection>:
$$
\left(f-\sum_ja_jN_j,N_i\right)_{L_2}=0,\qquad 1\le i\le n.
$$
Dividing each equation by $c_i$ yields
$$
Ga=d,\qquad G_{ij}=(M_i,N_j)_{L_2},\qquad d_i=(M_i,f)_{L_2}.
$$
This is the <mixed-normalization spline Gram matrix>. It need not be symmetric, but $G=D^{-1}H$ with $D=\operatorname{diag}(c_i)$ and $H_{ij}=(N_i,N_j)_{L_2}$. The ordinary <Gram matrix> $H$ is positive definite because the <splines> are linearly independent, so $G$ is invertible and $a=G^{-1}d$.
Since each $M_i$ is nonnegative and has integral one,
$$
|d_i|\le\int_0^1M_i(t)|f(t)|dt\le\|f\|_\infty,
\qquad\|d\|_{\ell_\infty}\le\|f\|_\infty.
$$
The induced <matrix> <operator norm> is the maximum absolute row sum, so
$$
\|a\|_{\ell_\infty}\le\|G^{-1}\|_{\ell_\infty\to\ell_\infty}\|f\|_\infty.
$$
Finally, nonnegativity and the <subpartition of unity for B-splines> give
$$
|P_{\mathcal S}f(t)|
=\left|\sum_ja_jN_j(t)\right|
\le\|a\|_{\ell_\infty}\sum_jN_j(t)
\le\|a\|_{\ell_\infty}.
$$
Taking the supremum over $t$ and then over $\|f\|_\infty\le1$ proves the <maximum-norm bound for spline projection>:
$$
\boxed{\|P_{\mathcal S}\|_\infty\le\|G^{-1}\|_{\ell_\infty\to\ell_\infty}.}
$$
The formulas also establish linearity and boundedness of the projection on continuous inputs. Its range is exactly the finite-dimensional <spline> space $\mathcal S$, since it fixes every <spline> in that space. Thus, when the <splines> are continuous, it is an operator $C[0,1]\to C[0,1]$ with range $\mathcal S$; it is not onto all of $C[0,1]$ as the printed wording says. For example, $e^t$ cannot belong to a fixed finite-degree <spline> space because it is not a <polynomial> on any nonempty knot interval. The missing $M_i$ normalization and this range correction are needed for a precise reading of the result.
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