= Solution
Construct an <economical two-state quantum cloner> by moving both actual outputs slightly toward one another relative to the two desired copies. Set $|a\rangle=|00\rangle$ and $|b\rangle=|++\rangle$. Their overlap is $1/2$, so their sum and difference yield the <orthonormal> vectors
$$
|u\rangle=\frac{|a\rangle+|b\rangle}{\sqrt3},\qquad
|v\rangle=|a\rangle-|b\rangle.
$$
The desired outputs have the symmetric expressions
$$
|a\rangle=\cos\frac\pi6|u\rangle+\sin\frac\pi6|v\rangle,
\qquad
|b\rangle=\cos\frac\pi6|u\rangle-\sin\frac\pi6|v\rangle.
$$
Choose the actual outputs to be
$$
|c_0\rangle=\cos\frac\pi8|u\rangle+\sin\frac\pi8|v\rangle,
\qquad
|c_+\rangle=\cos\frac\pi8|u\rangle-\sin\frac\pi8|v\rangle.
$$
They are normalized and satisfy
$$
\langle c_0|c_+\rangle=\cos\frac\pi4=\frac1{\sqrt2}
=\langle00|+0\rangle.
$$
Consequently the prescription $|00\rangle\mapsto|c_0\rangle$, $|+0\rangle\mapsto|c_+\rangle$ preserves the <Gram matrix> of the two inputs and defines a <linear isometry> on their span. The <unitary extension of a finite-dimensional isometry> gives a <unitary operator> on the same four-dimensional, two-<qubit> <Hilbert space>; no extra workspace is required.
For an explicit full-space construction, define $c=\cos(\pi/8)$, $s=\sin(\pi/8)$ and the <orthonormal basis>
$$
|p\rangle=c|00\rangle+s|10\rangle,\qquad
|m\rangle=s|00\rangle-c|10\rangle,\qquad |01\rangle,\ |11\rangle.
$$
Choose two further <orthonormal> vectors perpendicular to $u,v$,
$$
|w\rangle=\frac{|01\rangle-|10\rangle}{\sqrt2},\qquad
|t\rangle=\frac{|01\rangle+|10\rangle-2|11\rangle}{\sqrt6}.
$$
Then
$$
U=|u\rangle\langle p|+|v\rangle\langle m|
+|w\rangle\langle01|+|t\rangle\langle11|
$$
maps one <orthonormal basis> to another and is therefore <unitary>. Since $|00\rangle=c|p\rangle+s|m\rangle$ and $|+0\rangle=c|p\rangle-s|m\rangle$, it has exactly the required two outputs.
Finally,
$$
\langle a|c_0\rangle=\langle b|c_+\rangle
=\cos\frac\pi6\cos\frac\pi8+\sin\frac\pi6\sin\frac\pi8
=\cos\frac\pi{24}.
$$
Both inputs therefore have the same <global cloning fidelity>, and so does their average:
$$
\boxed{F_0=F_+=\overline F=\cos^2\frac\pi{24}\approx0.9829629131.}
$$
This proves the requested existence with a two-<qubit> <unitary gate>.
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