1. State the overall task: verify and correct the six numbered physics/math computations supplied by the user, show intermediate work, and give clear final values. 1. Problem 1: Verify the reported tension $T_1$. 1.1 The user reported $T_1=110.4990$ and rounded $T_1\approx110.5\ \mathrm{N}$. 1.2 Rounding step: $110.4990\approx110.5$ to three significant figures since the fourth digit is 4, so rounding yields 110.5. 1.3 Final (rounded) result: $T_1=110.5\ \mathrm{N}$. 2. Problem 2: Parse and check the listed algebraic expressions and the equation for $T$. 2.1 Evaluate the arithmetic expression $1-\sqrt{17}^2+7^2$. 2.2 Compute $\sqrt{17}^2=17$ and $7^2=49$, so $1-17+49=33$. 2.3 The user also wrote $27\sin\theta=29.61$ which gives $\sin\theta=\dfrac{29.61}{27}=1.0967$. 2.4 Since $\sin\theta>1$ this is not physically possible for a real angle; there is a data or algebra error in that step. 2.5 Solve the linear equation the user gave in the form $T-118\,\dfrac{\sqrt{334}}{7}=388.776$ for $T$. 2.6 Rearrangement gives $T=388.776+118\dfrac{\sqrt{334}}{7}$. 2.7 Numerical evaluation: $\sqrt{334}\approx18.276$ so $118\times18.276\approx2156.568$ and dividing by $7$ gives $\dfrac{2156.568}{7}\approx308.081$. 2.8 Adding the constant gives $T\approx388.776+308.081\approx696.857$. 2.9 Final (consistent) result for that equation: $T\approx696.86\ \mathrm{N}$. 3. Problem 3: Compute the rope tension for a mass on a slope with given acceleration. 3.1 Given: $m=51.96\ \mathrm{kg}$, slope angle $\theta=13^{\circ}36'=13+\dfrac{36}{60}=13.6^{\circ}$, $a=0.264\ \mathrm{m/s^2}$, and $g=9.81\ \mathrm{m/s^2}$. 3.2 Compute $\sin\theta$ with $\theta=13.6^{\circ}$: $\sin(13.6^{\circ})\approx0.23517$. 3.3 Compute $g\sin\theta=9.81\times0.23517\approx2.3075\ \mathrm{m/s^2}$. 3.4 Add the acceleration: $g\sin\theta+a\approx2.3075+0.264=2.5715\ \mathrm{m/s^2}$. 3.5 Rope force $F_{rope}=m\bigl(g\sin\theta+a\bigr)=51.96\times2.5715\approx133.68\ \mathrm{N}$. 3.6 Final result: $F_{rope}\approx133.7\ \mathrm{N}$. 4. Problem 4: Compute acceleration and tension $T_2$ for three masses. 4.1 Given (as interpreted): total applied force or net numerator $=50.05$ and masses $m_1=14.79$, $m_2=34.65$, $m_3=35.66$ (units kg implied). 4.2 Sum masses $m_1+m_2+m_3=14.79+34.65+35.66=85.10\ \mathrm{kg}$. 4.3 Acceleration $a=\dfrac{50.05}{85.10}\approx0.5883\ \mathrm{m/s^2}$. 4.4 Compute $T_2=(m_1+m_2)\,a=(14.79+34.65)\times0.5883=49.44\times0.5883\approx29.09\ \mathrm{N}$. 4.5 Final results: $a\approx0.5883\ \mathrm{m/s^2}$ and $T_2\approx29.09\ \mathrm{N}$. 5. Problem 5: Clarify frictional formulas and correct numeric friction value. 5.1 The user mixed several symbolic relations; the standard relations are: normal force $N$ and frictional force $f_k=\mu_k N$ for kinetic friction and $f_s\le\mu_s N$ for static. 5.2 If the user provided $N=63.9\ \mathrm{N}$ and $\mu_k=0.20$, then $f_k=\mu_k N=0.20\times63.9=12.78\ \mathrm{N}$. 5.3 If an additional term $0.57725$ is to be added (as the user wrote), the total becomes $12.78+0.57725=13.35725\ \mathrm{N}$. 5.4 The numeric value $109.81$ in the user text does not follow from the shown arithmetic; using the stated coefficients the consistent kinetic friction value is $\approx13.36\ \mathrm{N}$. 6. Problem 6: Force applied at an angle with friction; derive correct formula and compute numeric value. 6.1 Setup: horizontal component $F\sin\theta$ (user chose that convention) must balance kinetic friction $f_k=\mu_k N$ where $N=mg+F\cos\theta$ if $F$ has an upward component reducing normal force. 6.2 Equation: $F\sin\theta=\mu_k\bigl(mg+F\cos\theta\bigr)$. 6.3 Solve for $F$: $F\sin\theta-\mu_k F\cos\theta=\mu_k mg$ so $F\bigl(\sin\theta-\mu_k\cos\theta\bigr)=\mu_k mg$. 6.4 Therefore $F=\dfrac{\mu_k mg}{\sin\theta-\mu_k\cos\theta}$ if the sign convention has $F\sin\theta$ to the right and $F\cos\theta$ upward reducing $N$; equivalently if the geometry uses $F\sin\theta$ on the left as the user wrote, use the same sign convention shown here. 6.5 Using the user numbers interpreted as $m=0.72\ \mathrm{kg}$, $\mu_k=0.29$, $g=9.81$, $\sin\theta=0.5804$, $\cos\theta=0.80737$ gives $mg=0.72\times9.81=7.0632$. 6.6 Evaluate numerator $\mu_k mg=0.29\times7.0632\approx2.04853$ and denominator $\sin\theta-\mu_k\cos\theta=0.5804-0.29\times0.80737=0.5804-0.23414\approx0.34626$. 6.7 Thus $F\approx\dfrac{2.04853}{0.34626}\approx5.915\ \mathrm{N}$ under that sign convention. 6.8 If instead the correct equilibrium is $F\sin\theta=\mu_k(mg+F\cos\theta)$ with the algebra arranged as $F(\sin\theta+\mu_k\cos\theta)=\mu_k mg$ then 6.9 $F=\dfrac{\mu_k mg}{\sin\theta+\mu_k\cos\theta}=\dfrac{2.04853}{0.5804+0.23414}=\dfrac{2.04853}{0.81454}\approx2.515\ \mathrm{N}$. 6.10 Conclusion: which numeric value is correct depends on the sign convention for components and whether $F$'s vertical component increases or decreases the normal force; with the common assumption that $F$ has an upward component reducing $N$ the denominator is $\sin\theta-\mu_k\cos\theta$ and $F\approx5.92\ \mathrm{N}$, while with the alternate algebraic arrangement the consistent positive-denominator form gives $F\approx2.52\ \mathrm{N}$. Check the problem diagram to pick the correct sign convention. Final summary of corrected numerical answers (bracketed): 1. $T_1=110.5\ \mathrm{N}$. 2. Arithmetic: $1-\sqrt{17}^2+7^2=33$; inconsistent $\sin\theta$ noted; from $T-118\dfrac{\sqrt{334}}{7}=388.776$ we get $T\approx696.86\ \mathrm{N}$. 3. $F_{rope}\approx133.7\ \mathrm{N}$. 4. $a\approx0.5883\ \mathrm{m/s^2}$ and $T_2\approx29.09\ \mathrm{N}$. 5. Corrected kinetic friction using $\mu_k=0.20$ and $N=63.9\ \mathrm{N}$ is $f_k\approx13.36\ \mathrm{N}$; the listed 109.81 is not supported by the given arithmetic. 6. Depending on sign convention $F\approx5.92\ \mathrm{N}$ or, with the alternate algebraic arrangement, $F\approx2.52\ \mathrm{N}$; inspect the diagram to choose the correct sign convention.