Internal energy calculation
Your calculation is correct. The standardized definition of the change in internal energy $U$ for a closed thermodynamic system is

$$\Delta U=Q+W$$

where $Q$ is amount of heat transferred to the system and $W$ is work done on the system (provided that no chemical reactions occur). Therefore, heat transferred to the system is assigned a positive sign in the equation
$$Q=750\ \mathrm J$$
whereas work done by the system on the surroundings during the expansion of the liquid is assigned a negative sign
$$W=-200\ \mathrm J$$
Thus, the change in internal energy is
$$\begin{align}
\Delta U&amp;=Q+W\\
&amp;=750\ \mathrm J-200\ \mathrm J\\
&amp;=550\ \mathrm J\\
\end{align}$$

However, the question is a bit flawed since the given values are not typical for a liquid. By way of comparison, realistic values for water are shown in the following table.

$$\textbf{Water (liquid)}\\
\begin{array}{lllll}
\hline
\text{Quantity} &amp; \text{Symbol} &amp; \text{Initial value (0)} &amp; \text{Final value (1)} &amp; \text{Change}\ (\Delta) \\
\hline
\text{Amount of substance} &amp; n &amp; 1.00000\ \mathrm{mol} &amp; 1.00000\ \mathrm{mol} &amp; 0\\
\text{Volume} &amp; V &amp; 18.0476\ \mathrm{ml} &amp; 18.0938\ \mathrm{ml} &amp; 0.0462\ \mathrm{ml} \\
 &amp; &amp; 1.80476\times10^{-5}\ \mathrm{m^3} &amp; 1.80938\times10^{-5}\ \mathrm{m^3} &amp; 4.62\times10^{-8}\ \mathrm{m^3} \\
\text{Pressure} &amp; p &amp; 1.00000\ \mathrm{bar} &amp; 1.00000\ \mathrm{bar} &amp; 0 \\
 &amp; &amp; 100\,000\ \mathrm{Pa} &amp; 100\,000\ \mathrm{Pa} &amp; 0 \\
\text{Temperature} &amp; T &amp; 20.0000\ \mathrm{^\circ C} &amp; 29.9560\ \mathrm{^\circ C} &amp; 9.9560\ \mathrm{^\circ C} \\
 &amp; &amp; 293.1500\ \mathrm{K} &amp; 303.1060\ \mathrm{K} &amp; 9.9560\ \mathrm{K} \\
\text{Internal energy} &amp; U &amp; 1\,511.59\ \mathrm{J} &amp; 2\,261.58\ \mathrm{J} &amp; 749.99\ \mathrm{J} \\
\text{Enthalpy} &amp; H &amp; 1\,513.39\ \mathrm{J} &amp; 2\,263.39\ \mathrm{J} &amp; 750.00\ \mathrm{J} \\
\hline
\end{array}$$

When $1\ \mathrm{mol}$ of water with an initial temperature of $T_0=20\ \mathrm{^\circ C}$ is heated with $\Delta H=Q=750\ \mathrm J$ at a constant pressure of $p=1\ \mathrm{bar}$, the resulting expansion is actually only
$$\begin{align}
\Delta V&amp;=V_1-V_0\\
&amp;=18.0938\ \mathrm{ml}-18.0476\ \mathrm{ml}\\
&amp;=0.0462\ \mathrm{ml}\\
&amp;=4.62\times10^{-8}\ \mathrm{m^3}
\end{align}$$

The corresponding pressure-volume work is
$$\begin{align}
W&amp;=p\Delta V\\
&amp;=100\,000\ \mathrm{Pa}\times4.62\times10^{-8}\ \mathrm{m^3}\\
&amp;=0.00462\ \mathrm J
\end{align}$$
which is clearly below the value given in the question $(W=200\ \mathrm J)$.

The values given in the question are appropriate for a gas. For example, realistic values for nitrogen are shown in the following table.

$$\textbf{Nitrogen (gas)}\\
\begin{array}{lllll}
\hline
\text{Quantity} &amp; \text{Symbol} &amp; \text{Initial value (0)} &amp; \text{Final value (1)} &amp; \text{Change}\ (\Delta) \\
\hline
\text{Amount of substance} &amp; n &amp; 1.00000\ \mathrm{mol} &amp; 1.00000\ \mathrm{mol} &amp; 0\\
\text{Volume} &amp; V &amp; 24.3681\ \mathrm{l} &amp; 26.5104\ \mathrm{l} &amp; 2.1423\ \mathrm{l} \\
 &amp; &amp; 0.0243681\ \mathrm{m^3} &amp; 0.0265104\ \mathrm{m^3} &amp; 0.0021423\ \mathrm{m^3} \\
\text{Pressure} &amp; p &amp; 1.00000\ \mathrm{bar} &amp; 1.00000\ \mathrm{bar} &amp; 0 \\
 &amp; &amp; 100\,000\ \mathrm{Pa} &amp; 100\,000\ \mathrm{Pa} &amp; 0 \\
\text{Temperature} &amp; T &amp; 20.0000\ \mathrm{^\circ C} &amp; 45.7088\ \mathrm{^\circ C} &amp; 25.7088\ \mathrm{^\circ C} \\
 &amp; &amp; 293.1500\ \mathrm{K} &amp; 318.8588\ \mathrm{K} &amp; 25.7088\ \mathrm{K} \\
\text{Internal energy} &amp; U &amp; 6\,081.06\ \mathrm{J} &amp; 6\,616.83\ \mathrm{J} &amp; 535.77\ \mathrm{J} \\
\text{Enthalpy} &amp; H &amp; 8\,517.87\ \mathrm{J} &amp; 9\,267.87\ \mathrm{J} &amp; 750.00\ \mathrm{J} \\
\hline
\end{array}$$

When $1\ \mathrm{mol}$ of nitrogen with an initial temperature of $T_0=20\ \mathrm{^\circ C}$ is heated with $\Delta H=Q=750\ \mathrm J$ at a constant pressure of $p=1\ \mathrm{bar}$, the resulting pressure-volume work is

$$\begin{align}
W&amp;=p\Delta V\\
&amp;=100\,000\ \mathrm{Pa}\times0.0021423\ \mathrm{m^3}\\
&amp;=214.23\ \mathrm{J}
\end{align}$$
The corresponding enthalpy balance
$$\begin{align}
\Delta H&amp;=\Delta U+W\\
750.00\ \mathrm{J}&amp;=535.77\ \mathrm{J}+214.23\ \mathrm{J}
\end{align}$$
is quite similar to the values of the question $(\Delta H=Q=750\ \mathrm J,$ $\Delta U=550\ \mathrm J,$ and $W=200\ \mathrm{J}).$