= Solution
For the <unsigned interpretation> of the <binary string>, the set positions have weights $2^{15},2^{13},2^{12},2^{11},2^{10},2^8,2^6,2^5,2^4$. Thus
$$
\boxed{u=2^{15}+2^{13}+2^{12}+2^{11}+2^{10}+2^8+2^6+2^5+2^4=48496.}
$$
In 16-<bit> <two's complement>, the <most significant bit> is set, so subtract $2^{16}$ from that <unsigned interpretation>:
$$
\boxed{z=48496-65536=-17040.}
$$
For a <half-precision floating-point number>, split the fields at their actual boundaries rather than the printed groups of four:
$$
1\;01111\;0101110000.
$$
The sign is negative, the true <exponent> is $15-15=0$, and the fraction is $2^{-2}+2^{-4}+2^{-5}+2^{-6}=23/64$. It is a normal finite value, so
$$
\boxed{x=-\left(1+2^{-2}+2^{-4}+2^{-5}+2^{-6}\right)2^0=-\frac{87}{64}=-1.359375.}
$$
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