= Solution
Apply the <splitting principle for complex vector bundles> simultaneously to $E$ and $F$. This can be implemented by successive projective bundle pullbacks; the projective bundle theorem makes their composite pullback injective on integral <cohomology>. After pullback write $E=L_1\oplus L_2$ and $F=M_1\oplus M_2$, with <Chern roots> $x_1,x_2$ and $y_1,y_2$. Let
$$
A=x_1+x_2=c_1(E),\quad C=x_1x_2=c_2(E),\qquad
B=y_1+y_2=c_1(F),\quad D=y_1y_2=c_2(F).
$$
Since the <First Chern class> of a tensor product of <complex line bundles> is the sum of their first <Chern classes>, the four roots of $E\otimes F$ are $x_i+y_j$, $1\leq i,j\leq2$.
Compute their six pairwise products integrally. The two pairs within a fixed $x_i$ row contribute
$$
\sum_{i=1}^2(x_i+y_1)(x_i+y_2)=x_1^2+x_2^2+AB+2D=A^2-2C+AB+2D.
$$
The four pairs between the two rows contribute
$$
\sum_{j,k=1}^2(x_1+y_j)(x_2+y_k)=4C+2AB+B^2.
$$
Their sum is the second elementary symmetric <polynomial> of the tensor roots. Injectivity of the pullback returns the resulting identity to $X$, giving
$$
\boxed{c_2(E\otimes F)=c_1(E)^2+3c_1(E)c_1(F)+c_1(F)^2+2c_2(E)+2c_2(F).}
$$
Products here are <cup products> in integral <cohomology>. This <second Chern class of a tensor product of rank-two bundles> formula uses no division by two and therefore holds without assuming the integral <cohomology> of $X$ is torsion-free.
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