= Solution
{c}
Let $\beta(q)$ be the centered nondimensional cubic <B-spline>, even and supported on $[-2,2]$. The outer cubic must be $b(2-|q|)^3$ to make the function and its first two <derivatives> vanish at the support edge. Write its inner part as $A+Bq^2+C|q|^3$. Matching the value and first two <derivatives> at $|q|=1$ gives $A=4b$, $B=-6b$, $C=3b$. Its nonzero knot values are then $4b,b,b$, so requiring their sum to be one fixes $b=1/6$. Thus
$$
\boxed{\beta(q)=
\begin{cases}
(4-6q^2+3|q|^3)/6,&|q|\leq1,\\
(2-|q|)^3/6,&1\leq|q|\leq2,\\
0,&|q|\geq2.
\end{cases}}
$$
In particular $\beta(0)=2/3$, $\beta(\pm1)=1/6$.
With $h=1/m$, represent the <spline> on $[0,1]$ as $s(x)=\sum_{j=-1}^{m+1}c_j\beta(x/h-j)$. There are $m+3$ unknown coefficients. <interpolation> at $x_i=ih$, $0\leq i\leq m$, gives
$$
c_{i-1}+4c_i+c_{i+1}=6\sqrt{1-ih}.
$$
The natural endpoint conditions, using $\beta''(0)=-2$ and $\beta''(\pm1)=1$, give
$$
c_{-1}-2c_0+c_1=0,\qquad
c_{m-1}-2c_m+c_{m+1}=0.
$$
These $m+3$ equations are the required <matrix> system: the $m+1$ <interpolation> rows have coefficients $(1,4,1)$ at consecutive positions and right sides $6u(x_i)$; the two endpoint rows have $(1,-2,1)$ and zero right sides. No solution of that system is required. The scaling uses $\beta(x/h-j)$, without an extra $1/h$, because normalization is by knot values rather than by integral.
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