It looks as though the first force is
[tex]\vec F_1 = (\vec\imath-\vec\jmath)\,\mathrm N[/tex]
The second force acts in the direction of [tex]4\vec\imath+3\,\vec\jmath[/tex]. Normalize this direction vector to make it have length 1:
[tex]\left\|4\,\vec\imath+3\,\vec\jmath\right\| = \sqrt{4^2+3^2} = \sqrt{25} = 5 \\\\ \implies \left\|\dfrac{4\,\vec\imath+3\,\vec\jmath}5\right\| = 1[/tex]
Then the second force is
[tex]\vec F_2 = (10\,\mathrm N)\dfrac{4\,\vec\imath+3\,\vec\jmath}5 = (8\,\vec\imath+6\,\vec\jmath)\,\mathrm N[/tex]
The resultant force is
[tex]\vec F_1 + \vec F_2 = (9\,\vec\imath+5\,\vec\jmath)\,\mathrm N[/tex]
Assuming the particle moves 2 m in the same direction as the resultant force, the work perfomed on the particle by it is
[tex]\|\vec F_1 + \vec F_2\| (2\,\mathrm m) = 2\sqrt{9^2+5^2}\,\mathrm{Nm} \approx \boxed{20.6\,\mathrm J}[/tex]
