= Solution
For $0<t<1$, split the <Green's function> integral where the absolute value changes sign:
$$
x(t)=-\frac14\left[e^{-2t}\int_0^t\tau e^{2\tau}\,d\tau+e^{2t}\int_t^1\tau e^{-2\tau}\,d\tau\right].
$$
The two terms inside the bracket are respectively
$$
\frac t2-\frac14+\frac14e^{-2t},\qquad\frac t2+\frac14-\frac34e^{2t-2}.
$$
Adding them yields
$$
\boxed{x(t)=-\frac t4-\frac1{16}e^{-2t}+\frac3{16}e^{2t-2},\qquad0<t<1.}
$$
The affine term gives $x''-4x=t$, while both exponential terms are homogeneous solutions. The values at the joins are $x(0)=(3e^{-2}-1)/16$ and $x(1)=-(1+e^{-2})/16$, agreeing with parts (i) and (ii). Their first <derivatives> also match: $x'(0)=(3e^{-2}-1)/8$ and $x'(1)=(1+e^{-2})/8$. Thus the pieces introduce no unwanted <Dirac delta distributions>.
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